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32
where o is the (fundamental
Systems
frequency) power factor
The DC power must match the AC
power ignoring t
get Verter T'hus, We
Pac Pde =Vala v3E,,11 coso
=
Substituting for V and 11 from equations (2.3) and (2.6) in the above
(29
equation we
obtain
cOs = cos a
The reactive power
requirements are increased is (2.10
When a 90° . the
= increased from O (or reduced from
as a
power factor is zero and only reactive 180°
2.2.2 power is consumed.
Choice of Converter
mentioned earlier that GraetzConfiguration
for any Pulse
We Number
harmonic is sixth. bridge a six pulse converter for which
is
Correspondingly,
harmonic content in the AC lowest AC current harmonics the lowest DC
are 5th and 7th, To
voltage
In general, it can current and DC voltage, it is desirable to use higher reduce the
be stated that the
order characteristic harmonics (under ideal pulse numbers
conditions) are of the
hde np, hae =
nptl
where 'n' is an
integer. 'p is the
pulse number. (2.11)
There are several
have in addition configurations for a
converter of
phase full wave
to the Graetz specified pulse number. For p 6, we
bridge. six-phase diametral
a
=
converters, connection, cascade of three
interphase transformer etc. cascade of two three phase converter, single
It is convenient
to consider
parallel connection with
of basic
a
valve a'p pulse converter made of
switch is (commutation) of q valves or up series and
a
becomes positive). The
thyristor valve whosegroup
firing can be shown in Fig. 2.7.connections
switches as
parallel
voltage delayed (from the instant Here. the
windings. sources
Neglecting overlap, only are
actually obtained from when the valve
voltage
one valve conducts in the converter
a
commutation group of transformer
q' valves.
Va
e e =
Em cos wt
e2 =
Em COs (ot-
Em Cos (ot- 2r (q-1)
Fig. 2.7: A valve
group with 'q' valves
If the
converter is made up or a matix or S
Valve
parallel, then, groups in
series and ''
valye eroups in
P =qrs
(2.12)
Line Commutated and Voltage Source uonverters 33
Fig. 2.8: Converter made up of series and parallel connection of communication groups
See Fig. 2.8 for the converter configuration. In general, there are
p transformer windings.
It will be shown later that sometimes windings can be combined (in particular for
q 3, r = 1, s = 2).
Valve rating
The valve voltage rating is specified in terms of peak inverse
voltage (PIV) it has to withstand.
The ratio of PIV to the average DC voltage is an index of the valve utilization. The
average
maximum DC voltage across the converter is given by
ao89E, cosot do
Esin
(2.13)
The peak inverse voltage (PIV) across a valve can be obtained as follows:
If'd is even, then the maximum inverse voltage occurs when the valve
with a phase
displacement of t radian (180) is conducting and this is given by
PIV 2Em
If 'g' is odd, maximum inverse voltage Ccurs when the
valve with a phase shift of
T Tuq is conducting. In this case,
PIV = 2E cos
24 (2.14)
The valve utilization factor is given by
HVDC Poer Transmissio ems
34
PIV 2t
for q even
Vdo sq Sin
for q odd
sq sin
24
Table 2.3 shows the valve utilization factor for different six pulse converter configurations
The best valve utilization is obtained for configurations 1 and 3.
Table 2.3: Valve Utilization Factor
PIV
SI. No. S
Vdo
2 1 3 1.047
2 3 1 3.142
1 2 1.047
2 1 2.094
5 6 1 2.094
Transformer rating
The current rating of a valve (as well as transformer winding supplying it) is given by
(2.15)
where is the DC current which is assumed to be a constant. The transformer rating on the
valve side (in volt amperes) is given by
SP
p V2sqsin rVg
2 Jq sin (2.16)
The transformer utilization factor v
Su 18 only a function of 'q. The optimum value of q which
results in maximum utilization is equal to 3. It is a fortunate coincidence that the AC power
supply is 3 phase and the commutation group of 3 valves is easily arranged.
Source Converters 35
Line Commutated and Voltage
For q = 3
St 1.481 (2.17)
Vadodd
The transformer utilization can be improved further if two valve groups can share a single
transformer winding. In this case, the current (rms) rating of the winding can be increased by
a factor of 2 while decreasing the number of windings by a factor of 2. For this case,
Stu =1.047 (2.18)
Viodd
BA
a b 2
a t
4Z
Fig. 2.9: Six pulse converter
Fora 6 pulse converter, this can be easily arranged. The Graetz circuit shown in Fig. 2.1
s obtaned when the two winding (shown in Fig. 2.9) are combined into one. This is possible
sinee the terminals a and a'are at the same potential (with respect to the supply neutral point
n). Thus.it is shown that both from valve and transformeer utilization considerations, Graetz
circuit is the best circuit for a six pulse converter.
Twelve pulse converter
In HVDC transmission, the series conduction of converter groups has been preferred because of
the ease of control and protection as well as the requirements of high voltage rating. Thus a 12
pulse converter is obtained by the series connection of two bridges. The 30° phase displacement
between the two sets of source voltages is achieved by the transformer connections, Y/Y for feeding
one bridge and Y/A for feeding the second bridge.
The use of 12 pulse converter is preferable over the six pulse converter because of the reduced
iltering requirements. However, increase in pulse number beyond 12 is not practical because
the non-characteristic harmonics (which arise due to asymmetry in firing, imbalance in supply
etc.) are not eliminated.
36
HVDC Power Transmission
Systems
Example I
For a 12
pulse coonverter with q 4. s = =
3,r 1, calculate the maximum DC
=
power and transformer
ratings (valve windings) if PIV rating of the valve is V and the rms current
problem if q = 3, s = 4, r = 1. rating is I. Rework the
Solution
With even value of PIV
q. =
2E
12 V
ViosE,sin= T 2
in 4 32y -1.35V
I- I =2
Pdmax= Vdola =
2.7oVI
E, 12V
S= =:
2/
4.243V7
with q 3, s = 4, r = 1
TT
PIV 2E cos= 2E, cos 30 =
V3E
24
do ESin 12 V sin 60° = 6V
T S I n 60°
1.91V
I= ygrl =
v31, Pamax Vaol4 =
=1.91x V3vI =
3.31 VI
S =12-12V7 =
4.9VI
with only six windings on the valve side
4.9
Sw
S
2
VI =3.465VI
Line Commutated and Voltage Source Converters
37
dentical magnitude of
the AC voltages applied to the
defined as follows: two bridges. Assuming e4 ¬B and ec are
2
eA eas ELL Sin(ot+150)
B ebs 2
=
LL Sin(t +30°)
ec ees
2.
ELL sin(ot -90°)
We get the line line
to
voltages applied to the second bridge as
ebaDV3e = v2E, sin(ot -30)
(2.20)
Comparing the above equation with the Eq. (2.1) we note that the voltages
lag the voltages applied to the bridge 1 by 30°. applied to bridge 2
Assuming the identical delay angle for both the
bridges. valve 3 in the secon bridge will be fired 30° after firing the valve 3 in bridge 1.
Thus, in a 12 pulse converter. there are 12 intervals in a
cycle. each of duration 30°. In
each interval. 4 valves (2 each from the two
bridges) conduct. Table 2.4 shows the conducting
valves from each bridge for all the intervals (note that the
subscripts S and D denote the bridges
with star connected and delta connected windings
respectively).
Table 2.4: Bridge Voltages in Each Interval
Interval I (a) I b) II(a) I1(b) III(a I1I(bIV(a) IV(b)Va) VbVI (a) VI b)
Conducting
Valves 2, 3 3,4 4.5 .6 6.1 1.2
Bridge 1
ebcS ebaS ecaS ecbS eabS acS
Conducting
Valves 1, 2 2,3 3, 4 4, 5 5.6 6. 1 1.2
Bridge 2
a2 eacD ebcD ebaD caD ecbD abD acD
The DC voltage V, for a 12 pulse converter is the sum of the DC voltages for the individual
bridges. Thus,
Ua Ud1 +Ud2 (2.21)
For the interval I(a). the R.H.S. can be expressed as (see Table 2.1)
Ua ebes +eacD
= 2ELL Sin(ot +60°)- v2E;L sintot -270°)
2EL sin( ot + 60°) + cos ot|
E sin(ot + 75°) (2.22)
HVDC Power Transmission Systems
here E is defined as
E = 2E (2cos15)= v2E,, 1.93185) (2.23)
The average DC voltage (V) is given by
6
+30
Vde T
E,sin(ot + 75 )dot
- 2Vdo coso
(2.24)
where V=2E,, (see Eq. 2.3)
eA 'AS
00000 eaS
00000
eB B
BS
00000 00000 es Bridge
C cs 1 Vd1
00000 cS ecs
00000
AD
aD
BD
bD
CD
00000 Bridge
2
00000 CD N2
Fig. 2.10: A 12
pulse converter
The current igp flowing in the
(feeding bridge 2 in primary winding of the
Fig. 2.10) is given by star/delta connected
transforme
BD =3i
where i is the current (2.25
that i
flowng in the secondary
+i +ig 0.
=
we can show
that
winding that is coupled with phase B. Assumin
BD 1
3
(2.26
ot -k-67
-180° -120 120° 180a BS
60 60
1.154 d
0.577 d BD
2.154 d
1.577
0.577 l
pulse converter
Fig. 2.11: Waveforms of current in 12
shown in Fig. 2.11. In deriving the waveform of ian
The waveform of igs, iBD and ig a r e
120°.
it is to be noted that ibp lags ibs by 30° and iD lags iD by
fundamental component of the supply current in
value of the
It can be shown that the rms
a 12 pulse converter is given by
(2.27)
-2.
The harmonic current 1, is given by
h
where h = 12n t 1, n is an integer
is an even integer.
= 6m t 1, m
the
converter, it is obvious that
the harmonic currents ot a s i x pulse
with
In comparing
in the two transformers cancel each
other. This can be
fifth and seventh harmonics flowing
m is an odd integer.
generalized tothe case whenever
2.2.4 Effect of Finite Smoothing Reactor
reactance
which implies that the
is constant
In the previous sections, it
w a s assumed that l, reactor is finite,
reactor is very high (theoretically ifinite). If the smoothing
of the smoothing should be
the ripple in the DC current. The average DC current (1)
then we cannot neglect that the c u r r e n t
value, (corresponding to the peak value of the ripple) such
above a minimum
o
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