13

Constant reactive current control or AC voltage control may overcome some of the
problems of weak AC systems.
The power modulation techniques used to improve dynamic stability of power systems
will have to be modified in the presence of weak AC systems.

Six Pulse Converters

The conversion from AC to DC and vice-versa is done in HVDC converter stations by
using three phase bridge converters. The configuration of the bridge (also called Graetz circuit)
is a six pulse converter and the 12 pulse converter is composed of two bridges in series supplied
from two different (three-phase) transformers with voltages differing in phase by 30 o .

Pulse Number
The pulse number of a converter is defined as the number of pulsations (cycles of ripple)
of direct voltage per cycle of alternating voltage.
The conversion from AC to DC involves switching sequentially different sinusoidal
voltages onto the DC circuit.
 14

A valve can be treated as a controllable switch which can be turned ON at any instant,
provided the voltage across it is positive.
The output voltage Vd of the converter consists of a DC component and a ripple whose
frequency is determined by the pulse number

Choice of Converter Configuration

The configuration for a given pulse number is so chosen in such a way that the valve and
transformer are used to the maximum.

A converter configuration can be defined by the basic commutation group and the
number of such groups connected in series and parallel.
If there are ‘q’ valves in a basic commutation group and r of those are connected in
parallel and s of them in series then,
p=qrs
Note:
 15

A commutation group is defined as the group of valves in which only one (neglecting
overlap) conducts at a time.
Valve Rating:
The valve rating is specified in terms of Peak Inverse Voltage (PIV). The ratio of PIV to
average DC voltage is an index of valve utilization.
So, average maximum DC voltage across the converter is given by,

q
q
Vdo  s
2 
E m cos td (t )
q

q sq      sq 
s E m (sin t ) / q/ q  E m sin  sin    E m .2 sin
2 2  q  q  2 q

sq 
Vdo  E m sin ----- (1)
 q
If ‘q’ is even, then maximum inverse voltage occurs when the valve with a phase
displacement of 180o is conducting and is given by,
PIV = 2Em
If ‘q’ is odd, then maximum inverse voltage occurs when the valve with a phase shift of
π±(π/q) is conducting and is given by,
PIV = 2Em cos(π/2q)

The valve utilization factor is given by

PIV 2Em 2
For q even,  
Vdo sq  
E m sin s.q. sin
 q q

  
2 E m cos 2 . cos 2 . cos
PIV 2q 2q 2q
For q odd,   
Vdo sq    
E m sin sq. sin sq.2 cos sin
 q q 2q 2q
  2 
(Since sin2θ=2sinθcosθ and 2 cos sin  sin  sin )
2q 2q 2q q
 16

PIV 
 (For q odd)
Vdo 
sq. sin
2q

Transformer Rating:
The current rating of a valve is given by,
Id
Iv  ------ (2)
r q

where, Id is the DC current which is assumed to be constant.

The transformer rating on the valve side (in VA) is given by,
Em
S tv  p Iv
2
 17

From equations (1), (2) & p=qrs, we have

Vdo . Id
S tv  p .
 r q
2 .sq. sin
q

 Vdo I d
S tv  .
2 
q . sin
q

 S 
Transformer utilization factor  tv  is a function of q.
 Vdo I d 

As AC supply is three phase so, commutation group of three valves can be easily
arranged. So, for q = 3,
S tv 

Vdo I d 
(2 X 3) sin
3
S tv 

Vdo I d 6 sin 60 o
S tv
 1.48
Vdo I d
Transformer utilization can be improved if two valve groups can share single transformer
winding. In this case, the current rating of the winding can be increased by a factor of √2 while
decreasing the number of windings by a factor of 2.
 18

It is a 6-pulse converter consisting of two winding transformer where the transformer

utilization factor is increased when compared to three winding transformer.
The series conduction of converter groups has been preferred because of controlling and
protection as well as the requirements for high voltage ratings. So, a 12 pulse converter is
obtained by series connection of two bridges.
The 30o phase displacement between two sets of source voltages is achieved by
transformer connections Y-Y for one bridge and Y-∆ for the other bridge.
The use of a 12 pulse converter is preferable over the 6 pulse converter because of the
reduced filtering requirements.

Analysis of Graetz Circuit without overlap:

At any instant, two valves are conducting in the bridge, one from the upper commutation
group and the second from the lower commutation group. The firing of the next valve in a
particular group results in the turning OFF of the valve that is already conducting. The valves are
numbered in the sequence in which they are fired. Each valve conducts for 120 o and the interval
between consecutive firing pulse is 60 o in steady state.
The following assumptions are made to simplify the analysis
a. The DC current is constant.
b. The valves are modeled as ideal switches with zero impedance when ON and with
infinite impedance when OFF.
c. The AC voltages at the converter bus are sinusoidal and remain constant.
One period of the AC supply voltage can be divided into 6 intervals – each corresponding
to the conduction of a pair of valves. The DC voltage waveform repeats for each interval.
Assuming the firing of valve 3 is delayed by an angle α , the instantaneous DC voltage
Vd during the interval is given by
Vd = eb – ec = ebc for α ≤ ωt ≤ α+60o
Let eba  2 E LL sin t

then ebc  2E LL sin(t  60 o )

  60o
3
Average DC Voltage = Vd 
 

2 E LL sin(t  60 o )dt

3
 2 E LL [cos(  60 o  cos(  120 o )]

 19

3 2
Vd  E LL cos   1.35E LL cos 

Vd  Vdo cos  ------- (1)
The above equation indicates that for different values of α , V d is variable.
The range of α is 180 o and correspondingly Vd can vary from Vdo to –Vdo . Thus, the
same converter can act as a rectifier or inverter depending upon whether the DC voltage is
positive or negative.
DC Voltage Waveform:
The DC voltage waveform contains a ripple whose fundamental frequency is six times
the supply frequency. This can analyzed in Fourier series and contains harmonics of the order
h = np
where, p is the pulse number and n is an integer.
The rms value of the hth order harmonic in DC voltage is given by

2
Vh  Vdo [1  (h 2  1) sin 2  ]1 / 2
h 1
2

The waveforms of the direct voltage and calve voltage are shown for different values of
α.
 20

AC Current Waveform:
It is assumed that direct current has no ripple (or harmonics). The AC currents flowing
through the valve (secondary) and primary windings of the converter transformer contain
harmonics.

The waveform of the current in a valve winding is shown. The rms value of the
fundamental component of current is given by
 /3
1 2 6
I1 
2
I

 /3
d cos  .d 

I d ---- (2)

where as the rms value of the current is

2
I .I d
3
The harmonics contained in the current waveform are of the order given by
h  np  1
Where n is an integer, p is the pulse number. For a six pulse converter, the order of AC
harmonics is 5, 7, 11, 13 and higher order. These are filtered out by using tuned filters for each
one of the first four harmonics and a high pass filter for the remaining.
I1
The rms value of hth harmonic is given by Ih 
h
Power Factor:
The AC power supplied to the converter is given by
PAC  3E LL I1 cos 

Where cos  is the power factor.

The DC power must match the AC power ignoring the losses in the converter. Thus,
PAC  PDC  Vdo I d  3E LL I1 cos 

Substituting for Vdo and I1 from equations (1) and (2) in the above equation, we get
cos   cos 
 21

The reactive power requirements are increased as α is increased from zero (or reduced
from 180o ).

Analysis of Graetz Circuit with overlap

Due to the leakage inductance of the converter transformers and the impedance in the
supply network, the current in a valve cannot change suddenly and this commutation from one
valve to the next cannot be instantaneous. This is called overlap and its duration is measured by
the overlap (commutation) angle ‘μ’.
Each interval of the period of supply can be divided into two subintervals as shown in the
below timing diagram. In the first subinterval, three valves are conducting and in the second
subinterval, two valves are conducting which is based on the assumption that the overlap angle is
less than 60o .

There are three modes of the converter which are

i) Mode 1 – Two and three valve conduction (μ<60o )
ii) Mode 2 – Three valve conduction (μ=60o )
iii) Mode 3 – Three and four valve conduction (μ>60o )
i)Analysis of Two and Three Valve Conduction Mode:
The equivalent circuit for three valve conduction is shown below.
For this circuit,
 di di 
eb  ea  Lc  3  1 
 dt dt 
The LHS in the above equation is called the commutating emf whose value is given by
eb  ea  2 E LL sin t
Which is the voltage across valve 3 just before it starts conducting.
 22

Since, i1  I d  i3

We get,
di3
2 E LL sin t  2 Lc
dt
Solving the above equation, we get
i3 (t )  I s (cos   cos t ),  t    
Where,
2 E LL
Is 
2Lc

At ωt=α+μ , is = Id . This gives I d  I s [cos   cos(   )]

The average direct voltage can be obtained as

3 
    60
3
Vd  
   ec d (t )   (eb  ec )d (t )
 2   
3
 Vdo cos   2 E LL [cos   cos(   )]
2

3 2
Since, E LL  Vdo , we get

Vdo
Vd  [cos   cos(   )]
2
The value of [cos   cos(   )] can be substituted to get,

 I 
Vd  Vdo  cos   d   Vdo cos   Rc I d
 2I s 
Where,
3 3
Rc  Lc  Xc
 
Rc is called equivalent commutation resistance and the equivalent circuit for a bridge
converter is shown below.
 23

Inverter Equations:
For an inverter, advance angle β is given by
β=π-α
and use opposite polarity for the DC voltage with voltage rise opposite to the direction of
current. Thus,
 Vdoi
Vdi  [cos   cos(   )]
2
 Vdoi
 [cos(   )  cos(   )]
2
Vdoi
Vdi  [cos   cos  ]
2
Where, the extinction angle γ is defined as
γ = β-μ = π-α-μ
Similarly, it can be shown that
Vdi  Vdoi cos   Rci I d

 Vdoi cos   Rci I d

The subscript “i” refers to the inverter.
ii) Analysis of Three and Four Valve Conduction Mode:
The equivalent circuit for three and four valve conduction is shown below.
For, α ≤ ωt ≤ α+μ-60o
i1 = Is sin(ωt+60o ) + A
i6 = Id – i2 = Id – Is sinωt + C
Em 2
Where, I s   Is
Lc 3
The constant A can be determined from the initial condition
i1 (ωt=α) = Id = Is sin(α+60o ) + A
The constant C can be determined from the final condition
i6 (ωt=α+μ-60o ) = 0 = Id – Is sin(α+μ-60o ) + C = 0
 24

For, α+μ-60o ≤ ωt ≤ α+600

i1 = Is cosωt + B
The constant B can be determined from the continuity equation
i1 (ωt=α+μ=60o ) = Is sin(α+μ) + A = Is cos(α+μ-60o ) + B
Finally,
Is
Id  [cos(  30 o )  cos(    30 o )]
2
The expression for average direct voltage is given by
  60o
3 3
Vd  
    60 o 2
ec d (t )

Since ec = Em cosωt
33
Vd  Em [sin(  60 o )  sin(    60 o )]
 2
3
Vd  Vdo [cos(  30 o )  cos(    30 o )]
2
Finally
3 Id
Vd  Vdo [ 3 cos(  30 o )  ]  3.Vdo cos(  30 o )  3Rc I d
2 Is

Converter Bridge Characteristics

A) Rectifier: The rectifier has three modes of operation.
1) First mode: Two and three valve conduction mode ( μ < 60 o )
2) Second mode: Three valve conduction mode only for α < 30 o ( μ = 60o )
3) Third mode: Three and four valve conduction mode α ≥ 30 o ( 60o ≤ μ ≤ 120o )
As the DC current continues to increase, the converter operation changes over from
mode 1 to 2 and finally to mode 3.
The DC voltage continues to decrease until it reaches zero.
 25

For α ≥ 30o , mode 2 is bypassed.

For Modes 1 and 3, we have
Vd I
 cos   d
Vdo 2I s

Vd 3I
 3 cos(  30 o )  d
Vdo 2I s
The voltage and current characteristics are linear with different slopes in these cases.
For mode 2, μ = 60o , μ is constant, so the characteristics are elliptical and is given by
2 2
   
 V|   I| 
 d   d  1
 cos    sin  
   
 2   2 
Vd I
where, Vd|  and I d|  d
Vdo 2I s
B) Inverter:
The inverter characteristics are similar to the rectifier characteristics. However, the
operation as an inverter requires a minimum commutation margin angle during which the voltage
across the valve is negative. Hence the operating region of an inverter is different from that for a
rectifier.
So, the margin angle (ξ) has different relationship to γ depending on the range of
operation which are
First Range: β < 60o and ξ = γ
Second Range: 60o < β < 90o and ξ =60o – μ = γ-(β-60o )
Third Range: β > 90o and ξ = γ – 30o
In the inverter operation, it is necessary to maintain a certain minimum margin angle ξ o
which results in 3 sub-modes of the 1st mode which are

Mode 1
1(a) β < 60o for values of μ < (60o - ξo )
The characteristics are linear defined by
Vd| = cosγo – Id|
1(b) 60o < β < 90o for
μ = 60o – ξo = 60o – γo = constant
 26

The characteristics are elliptical.

1(c) 90 < β < 90o + ξo for values of μ in the range
o

60o – ξo ≤ μ ≤ 60o
The characteristics in this case are line and defined by
Vd| = cos( γo + 30o ) - Id|
Mode 2
For μ > 60o corresponding to β > 90o + γo
The characteristics again are linear but with a different slope and is defined by
Vd| = √3 cosγo - 3Id|
In the normal operation of the converter I d| is in the range of 0.08 to 0.1 .
Characteristics of a twelve pulse converter

As long as the AC voltages at the converter bus remain sinusoidal (with effective
filtering), the operation of one bridge is unaffected by the operation of the other bridge connected
in series. The region of rectifier operation can be divided into five modes as
Mode 1: 4 and 5 valve conduction
0 < μ < 30o
Mode 2: 5 and 6 valve conduction
30o < μ < 60o
Mode 3: 6 valve conduction
0 < α < 30o , μ = 60o
Mode 4: 6 and 7 valve conduction
60o < μ < 90o
 27

Mode 5: 7 and 8 valve conduction

90o < μ < 120o
The second mode is a continuation of the first and similarly fifth is a continuation of the
fourth.
The equivalent circuit of the twelve pulse converter is the series combination of the
equivalent circuits for the two bridges. This is because the two bridges are connected in series on
the DC side and in parallel on the AC side. The current waveforms in the primary winding of the
star/star and star/delta connected transformers and the line current injected into the converter bus
are shown.

Questions
1) What is the need for interconnection of systems? Explain the merits of connecting HVAC
systems by HVDC tie-lines?
2) (a) Discuss the different factors that favor HVDC transmission systems over EHVAC
transmission over long distances.
(b) What are the different HVDC links normally adopted?
3) (a) With the help of a neat schematic diagram of a typical HVDC converter station explain the
functions of various components available.
(b) What are the applications and merits of HVDC transmission system?
4) (a) Explain for what reasons as a system planner, you consider the applications of HVDC in
India?
 28

(b) Compare HVDC transmission system with AC system in all aspects.

5) For a 3-Φ, 6 pulse Graetz’s circuit, draw the timing diagram considering overlap angle is less
than 60o and without overlap for the following:
(a) Voltage across load
(b) Voltage across any two pair of conduction values
6) Explain the operation of a 12 pulse bridge rectifier with the help of circuit diagram, voltage &
current waveforms.
7) (a) Clearly explain how harmonics are produced and obtain the expression for rms value of the
fundamental component of the current.
(b) Obtain a relation between firing angle and power factor angle in a 3-Φ bridge rectifier.
8) Derive the expression for average DC voltages of a six pulse bridge converter, considering
gate control and source reactance.
9) What is the reason for using star-star and star-delta transformer configurations for 12 pulse
converter? Derive an equation for primary current using fourier analysis.