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MODULE II
SINGLE PHASE FULL WAVE CONTROLLED RECTIFIER (TWO QUADRANT
CONVERTERS)
There are two basic configurations of full wave controlled rectifiers. Their classification
is based on the type of SCR configuration employed. They are :
(1) Mid-point converter (2) Bridge converter
Single phase full wave Mid point converters (M2 connection) with R load
These converters are also referred to as two pulse converters as two triggering pulses
are to be generated during every cycle of the supply. These are used for rectifiers of low
ratings.
Figure shows a 2 pulse mid point converter circuit with resistive load. This type of
full wave rectifier circuit uses 2 SCRs connected to the centre-tapped secondary of a
transformer. During the +ve half cycle of the a.c supply ,i.e when terminal A of the
transformer is +ve w.r.to terminal B, or the secondary winding terminal A is +ve w.r.to N,
SCR1(T1) is forward biased and SCR2(T2)is reverse biased. When T1 is triggered at a firing
angle α, current would flow from terminal A through T1, the resistive load R and back to the
centre tap of the transformer (terminal N). this current path is also shown in figure. This
current continuous to flow up to angle Π when the line voltage reverse its polarity and SCR1
is turned off. Depending upon the value of α and the load circuit parameters, the conduction
angle of SCR1 may be any value between 0 and Π.
Full wave mid-point circuit with resistive load
Waveforms for M-2 configuration with resistive load
During the –ve half cycle of the ac supply,, the terminal B of the transformer is +ve
w.r.t N, T2 is forward biased. When T2 is fired at an angle (Π+α), current would flow from
terminal B, through T2, the resistive load and back to centre tap of the transformer. This
current continuous till angle 2Π, then T2 is turned off. Here it is assumed that both thyristors
are triggered with same firing angle, hence they share the load current equally. The voltage
and current waveforms are as shown in figure. It is clear from this figure that with purely
resistive load, the load current is always discontinuous.
Average D.C output voltage
E0 
E
1
Em sin t.d (t )  m (1  cos  )



1. Average load current I0
I0 
Eo Em

(1  cos  )
R R
2. RMS load voltage
1/ 2
1  2

Eor  Erm s    Em sin 2 t.d (t )
 

  -  sin 2 
E rms  E m 

4 
 2
1/ 2
3. RMS load current
I RMS 
Eor
R
Single phase full wave Mid point converters (M2 connection) with R L load
The circuit diagram of a single phase full wave converter using a centre tapped transformer is
shown in figure.(a) when terminal a is +ve w.r.t ‘n’ terminal ‘n’ is +ve w.r.t b. therefore van =
vnb or van = -vbn as ‘n’ is the mid point of secondary winding. Equivalent circuit of this
arrangement is shown in figure (b). it is assumed that load or output current is continuous and
turns ratio from primary to each secondary is unity.
Single phase full wave mid-point converter
(a) Circuit diagram (b) equivalent circuit (c) various voltage & current waveforms
Thyristors T1 and T2 are forward biased during +ve and –ve half cycles respectively
and triggered accordingly. Suppose T1 is already conducting. After ωt = 0, van is +ve, T1 is
then forward biased and when triggered at delay angle α, T1 gets turned on. At this firing
angle α, T1 gets turned on. At this firing angle supply voltage 2Vm sinα reverse biases T2, this
SCR is therefore turned off. Here T1 is the incoming SCR and T2 is the outgoing SCR. As the
incoming SCR is triggered, ac supply voltage applies reverse bias across the outgoing SCR
and turns it off. Load current is also transferred from outgoing SCR to incoming SCR. This
process of SCR turns off by natural reversal of ac supply voltage is called natural or
line commutation.
From the equivalent circuit, it is seen that if
van =Vm sinωt,
then vbn = -vnb = Vm sinωt
and vab = van + vnb = 2Vm sinωt
when ωt = α, T1 is triggered. SCR T2 is subjected to a reverse voltage vab = 2 V m sinα.
Current is transferred from T2 to T1 and as a result T2 is turned off. The magnitude of
reverse voltage across T2 can also be obtained by applying KVL.
i.e vT2 – vbn +van-vT1 =0
vT2 = vbn- van +vT1
with T1 conducting , vT1 =0. Therefore the voltage across T2 at the instant ωt =α is given by
vT2 = - Vm sinα - Vm sinα = Vm sinα
this show that T2 is reverse biased by this voltage 2 Vm sinα Vm sinα is turned off at ωt = α .
T1 conducts from α to Π+α . After ωt = Π, T1 is reverse biased but it will continue
conducting as the forward biased SCR T2 is not get gated. At ωt = Π+α , T2 is triggered, T1
is reverse biased by voltage of magnitude 2 Vm sinα, current is transferred from T1 to T2, T1
is therefore turned off.
At ωt =α, T2 is turned off and it remains reverse biased from ωt = α to Π. The turn off time
provided by this circuit to SCR T2 is therefore given by
tc 
 

sec
Thyristor T1 is turned off at ωt = Π+α and it is reverse biased from ωt = Π+α to ωt =2 Π.
Therefore this circuit provides a turn off time to thyristor T1 as
tc 
2  (   )


 

sec
Which is same as provided to T2.
The average value of output voltage is given by
V0 
2V
1  
Vm sin t.d (t )  m cos 

 

In this the SCR is subjected to a peak voltage of 2Vm. from the above, we can observe that
(i)
When commutation of SCR is desired, it must be reverse biased and the incoming
SCR must be forward biased.
(ii)
When incoming SCR is gated on, current is transferred from outgoing SCR to
incoming SCR
(iii)
The circuit turnoff time must be greater than SCR turn off time.
SINGLE-PHASE FULL WAVE BRIDGE CONVERTERS
Phase controlled single phase or three phase, full-wave converters are primarily of
three types; namely (1) uncontrolled converters,
(2) half controlled converters and
(3) fully-converters.
An uncontrolled converter or rectifier uses only diodes and the level of dc output
voltage cannot be controlled. A half-controlled converter or semi-converter uses a mixture of
diodes and thyristors and there is a limited control over the level of dc output voltage.A
fully-controlled converter or full converter uses thyristors only and there is a wider control
over the level of dc output voltage.
A semiconverter is one-quadrant converter. A one-quadrant converter has one polarity
of dc output voltage and current at its output terminals as shown in figure.
(a) One quadrant converter (b) two quadrant converter
A two-quadrant converter is one in which voltage polarity can reverse but current
direction cannot reverse because of the unidirectional nature of thyristors fig (b) shown
above.
Single phase full wave bridge converters ( B-2 Connection) with R load
A single phase full converter bridge using four SCRs is shown in figure.
Single phase full wave controlled bridge rectifier with R load
Figure shows a single phase fully controlled (i.e both half cycles are phase controlled )
bridge converter, supplying a resistive load. During the +ve half cycle thyristors T1 and T2
are forward biased. When T1 & T2 are triggered at ωt = α, they start conducting and
supplying the load current. At ωt=Π, current is zero, the supply voltage reverses and T1,T2
are turned off by natural commutation. In the –ve half cycle T3 & T4 conduct from ωt =Π+α
to ωt =2Π. At ωt =2Π, current is zero, the supply voltage reverses and T3 &T4 are turned off
by natural commutation. The above sequence of events is repeated in each cycle. The wave
forms of input, firing pulse, output voltage and load currents are also shown below.
Voltage and current waveform for single phase full-wave controlled rectifier
The expressions for Vo, Io Vrms are same as that of given in M -2 connection.
Fully controlled Bridge Circuit with R-L Load:
The single phase fully controlled bridge circuit with R-L load is shown in figure. Conduction
doesnot take place unitl the thyristors are fired and in order for current to flow, thyristors T1
and T2 must be fired together, as must T3 and T4 in the next half cycle. Inductance L is used
in the circuit to reduce the ripple. A large value of L will result in a continuous steady current
in the load. A small value of L will produce a discontinuous load current for large firing
angles. The wave forms are given in figure.
Fully controlled single phase bridge with R-L load
The voltage waveform at the dc terminals comprises a steady d.c component on to
which is superimposed an a.c ripple component, having a fundamental frequency equal to
twice that of a.c supply.
Single phase full wave bridge converters ( B-2 Connection) with R –L-E load
The figure (a) shows a single phase full converter bridge using 4 SCR and the load is
assumed to be RLE type. E is the load circuit emf. voltage E may be due to a battery in the
load circuit. or may be generated emf of a dc motor. SCR T1 & T2 is simultaneously
triggered and 180º later pair T3,T4 is gated together. When a is +ve w.r.t b, supply voltage
waveform is shown as vab in figure (b).
(a) Single phase full converter bridge with RLE load
When b is +ve w.r t ‘a’ supply voltage waveform is shown dotted as vba . obviously vab = vba. The current directions and voltage polarities in figure (a) are treated as +ve.
Fig (b) voltage and current wave forms for continuous load current
Load current io is assumed continuous over the working range.i.e load is alaways
connected to the ac voltage source through the thyristors. 0<α<ωt, T1,T2 are forward biased
through the already conducting SCRs T3,T4 and block the forward voltage. For continuous
current, T3,T4 conduct after ωt = 0 even though these are reverse biased. When forward
biased SCRs T1,T2 are triggered at ωt = α, they get turned on. As a result, supply voltage V m
sinα immediately appears across thyristors T3,T4 as a reverse bias, these are therefore turned
off by natural, or line commutation.at the same time load current io flowing through T3,T4 is
transferred to T1,T2 at ωt = α. When T1,T2 are gated at ωt = α, these SCRs will get
turned on only if Vm sinα > E. SCR T1,T2 conduct from ωt = α to Π +α. At ωt = Π +α,
forward biased SCRs T3,T4 are triggered. The supply voltage turns off T1, T2 by natural
commutation and the load current is transferred from T1,T2 to T3,T4.
Voltage across T1,T2 is shown as vT1,vT2, and that across T3,T4 as vT3,vT4.
Source current is is taken as +ve in the arrow direction. Hence source current is shown +ve
when T1,T2 are conducting and –ve when T3,T4 are conducting.
During α to Π, both vs and is are +ve, power flows from source to load. During Π to Π +α, vs
is –ve but is +ve, hence load returns some of its energy to the supply system. But the net
power flow is from ac source to dc load because
(Π –α) > α.
The average value of output voltage Vo is given by
V0 
2V
1  
Vm sin t.d (t )  m cos 

 

(1)
RMS value of output voltage for single phase M-2 or B-2 controlled converter can be
obtained as below.
1/ 2
 1   2 2

Vor  
Vm sin t.d (t )

 

 Vm 2 / 2  Vs
Vor  Vs
Equation (1) shows that if α > 90º, Vo is –ve. This is illustrated in fig ( c), where α is shown
greater than 90°. In this Vo is –ve. if the load circuit emf E is reversed, this source E will feed
power back to ac supply. This operation of full converter is known as inverter operation of
the converter. The full converter with firing angle delay greater than 90º is called line
commutated inverter. Such an operation is used in the regenerative breaking mode of a dc
motor in which case then E is counter emf of the dc motor.
Fig (c ). Voltage and current waveform for single phase full converter for α > 90°
During 0 to α, ac source voltage vs is +ve but ac source current is is –ve , power flows from dc
source to ac source. From α to Π, both vs & is are +ve hence power flows from ac source to dc
source. But the net power flow is from dc to ac source because (Π-α) < α.
In converter operation, the average value of output voltage V0 must be greater than load
circuit emf E. during inverter operation, load circuit emf when inverted to ac must be more
than ac supply voltage. In other words, dc source voltage E must be more than inverter
voltage V0 , only then power would flow from dc source to ac supply system. But in both
converter and inverter modes, thyristors must be forward biased and current through SCRs
must flow in the same direction as these are unidirectional devices. This is the reason output
Current io is shown +ve in fig(c). As before source current is +ve when T1,T2 are conducting.
The variation of voltage across thyristors T1,T2,T3 or T4 reveals that circuit turn-off
time for both converter and inverter operations is given by
tc 
 

sec
Advantages of single phase bridge converter over mid point converter
1. SCRs are subjected to a peak inverse voltage 2Vm in mid point converter and Vm in
bridge converter. Thus for the same voltage and current ratings of SCRs, power
handled by mid point configuration is about half of that handled by bridge converter.
2. In mid point converter, each secondary should be able to supply the load power. As
such the transformer rating in mid point converter is double the load rating.
From these above, we can conclude that bridge converter is preferred over mid point
configuration.
Single phase half controlled Bridge converter ( Semi converter)R load
A semi converter uses a mixture of diodes and thyristors and there is a limited control
over the level of dc output voltage. A semi converter is one quadrant type and it has
one polarity of dc output voltage and current at its output terminal and is always +ve.
It is also known as 2 pulse converter.
The figure shows half controlled rectifier with R load. This circuit consists of 2 SCRs
T1,T2 and 2 diodes D1,D2.
1 phase half wave controlled rectifier with R load
During positive half cycle of the ac supply, SCR T1 and diode D1 are forward
biased and when T1 is triggered at a firing angle ωt = α, the SCR T1, diode D1 comes
to the on state. Now the load current flows through P-T1- R- D1-N. during this period,
we can get output voltage and current are +ve.
At ωt = Π, the load voltage and current reaches to zero. Then SCR T1 and diode D1
comes to off state due to natural commutation.
During the negative half cycle of the supply T2 and D2 are forward biased. When
SCR T2 is triggered at a firing angle ωt = Π+α, the SCR T2 and diode D2 comes to on
state. Now the load current flows through the path N- T2- R- D2- P. During this
period we can get +ve output voltage and + ve output current.
At ωt = 2Π, the load voltage and load current reaches to zero then SCR T2 and diode
D2 comes off state due to natural commutation. During the period (Π+α to 2Π) SCR
T2 and diode D2 are conducting.
4. Average load voltage Vo
V0 
V
1
Vm sin t.d (t )  m (1  cos  )



5. Average load current I0
I0 
Vo Vm

(1  cos  )
R R
6. RMS load voltage
1/ 2
1  2

Vor  Vrm s    Vm sin 2 t.d (t )
 

  -  sin 2 
Vrms  Vm 

4 
 2
1/ 2
7. RMS load current
I RMS 
Vor
R
Single phase half controlled Bridge converter ( Semi converter)R –L load
The figure shows 2 pulse half controlled bridge rectifier with inductive load. Here the
inductance value should be large so that load current should be continuous. The
voltage and current waveforms as shown in figure.
1 phase half wave controlled rectifier with R-L load
During the +ve half cycle ( 0 to Π) SCR T1 and diode D1 are forward biased.
At ωt =α, SCR and diode D1 comes to the ON state. Now the current flows through
path P- T1-R-L-D1-N.
When the supply voltage reverses at ωt=Π, the diode D2 is forward biased since D1 is
already conducting. D2 comes to the ON state and the load current passes through D2
and T1. Now D1 in off stage due to supply voltage. Thus the load current freewheels
through the path L-D2-T1-R during the period Π to (Π+α).during this period output
voltage should be zero because of closed current path.
During –ve half cycle, at the instant (Π+α), a triggering pulse is applied to the forward
biased SCR T2,then T2 turned on. As T2 turned on , the supply voltage reverse biases
T1 and then it off by the natural commutation. During the period (Π+α) to 2Π, SCR
T2 and diode D2are conducting. Now the load current flows through the path N-T2R-L-D2-P. Thus we get +ve output voltage and current.
Now the supply voltage reverses at ωt = 2Π, the diode D1 forward biased since D2 is
already conducting. Then D1 comes to on state and load current passes through D1
and T2. The diode D2 is reverse biases due to supply voltage and it turns off. Thus
load current freewheels through the path L-D1-T2-R during the interval from 2Π to 2
Π+α. During this period output current is +ve but output voltage is zero due to closed
circuit. Here the conduction period of thyristors and diodes are equal, therefore this
circuit is called as symmetrical configuration.
Expression for output voltage
1. Average load voltage Vo
V0 
V
1
Vm sin t.d (t )  m (1  cos  )



2. RMS load voltage
1/ 2
1  2

Vor  Vrm s    Vm sin 2 t.d (t )
 

  -  sin 2 
Vrms  Vm 

4 
 2
1/ 2
As both the types of phase-controlled converters have been studied, the advantages of singlephase bridge converter over single-phase mid-point converter can now be stated:
i)
SCRs are subjected to a peak inverse voltage of 2Vm in mid-point converter and
Vm in full converter. Thus for the same voltage and current ratings of SCRs,
power handled by mid-point configuration is about half of that handled by bridge
configuration.
ii)
In mid-point converter, each secondary should be able to supply the load power.
As such the transformer rating in mid-point converter is double the load rating.
This however is not the case in single phase bridge converter.
It may thus be inferred from above that bridge configuration is preferred over mid-point
configuration. However, the choice between these two types depends primarily on cost of the
various components, available source voltage and the load voltage required. Mid-point
configuration is used in case the terminals on dc side have to be grounded.
Single-phase semiconverter with R-L-E Load
A single –phase semiconverter bridge with two thyristors and three diodes is shown
in fig (a).
Fig (a) Single phase semi converter bridge- power circuit diagram with RLE load
The two thyristors are T1,T2, the two diodes are D1,D2; the third diode connected
across load is freewheeling diode FD. The load is of RLE type as for the full converter
bridge. Various voltage and current waveforms for this converter as shown in fig (b), where
load current is assumed continuous over the working range.
(b) Voltage and current waveforms for continuous load current
After ωt=0,thyristor T1 is forward biased only when source voltage of Vmsinωt exceeds E.
Thus T1 is triggered at a firing delay of α .such that of vmsinα > E.With T1 on ,load gets
connected to source through T1 and D1.For the period ωt = of α to Π ,load current i0 flows
through RLE, D1 source and T1 and the load terminal voltage v0 is of the same wave shape
as the ac voltage source vs. Soon after ωt= Π,load voltage v0 tends to reverse as the ac voltage
changes polarity. Just as v0 tends to reverse at ωt=(Π+), FD gets forward biased and starts
conducting. The load or output current i0 is transferred from T1,D1 to FD.As SCR T1 is
reverse biased at ωt=(Π+), through FD,T1 is turned off at ωt=(Π+ ), The waveform of current
iT1 through thyristor T1 is shown in figure (b)
It flows from α to Π, 2Π+α to 3Π and so on for for an interval of (Π-α) radians. The load
terminals are short circuited through FD, therefore load or output voltage v0 is zero during Π
< ωt < (Π+α). After ωt=Π, during the negative half cycle ,T2 will be forward biased only
when source voltage is more than E. At ωt=Π+α, source voltage exceeds E, T2 is therefore
triggered. Soon after (Π+α), FD is reverse biased and is therefore turned off; load current now
shifts from FD to T2,D2.At ωt=2Π,FD is again forward biased and output current i0 is
positive from α toΠ when T1,D1 conduct and is negative from (Π+α) to 2Π when T2,D2
conduct.
During the interval α to Π,T1 and D1 conduct and ac source delivers energy to the load
circuit. This energy is partially stored in inductance L, partially stored as electric in load –
circuit emf E and partially dissipated as heat in R, during the freewheeling period Π to
(Π+α),energy stored in inductance is recovered and partially added to the energy stored in
load emf E. No energy is fed back to the source during freewheeling period. For semiconverter, the average output voltage v0 from the wave form ,
V0 
V
1
Vm sin t.d (t )  m (1  cos  )



And rms value of output voltage is
1/ 2
1  2

Vor  Vrm s    Vm sin 2 t.d (t )
 

  -  sin 2 
Vrms  Vm 

4 
 2
1/ 2
The circuit turn off time for the semi converter is
tc 
 

sec
PERFORMANCE PARAMETERS OF TWO PULSE CONVERTER
In general the instantaneous input current to a converter can be expressed in Fourier series as
under:

it   a0   an cos nt  bn sin nt 
n 1, 2,3
where a 0 
an 
1 2
 i(t )d (t )
2 0
1 2
 i(t ) cos nt d(t) and
 0
bn 
1 2
 i(t ) sin nt d(t)
 0
The performance parameters are now obtain ned first for single phase full convereter AND
THEN for semi converter.
Single phase full converter
From the waveform of full converter, the variation of input current, or source current is, from
α to (Π+α), from (Π+α) to (2Π+α) is continuous but not constant. Here is is assumed to be
ripple free with amplitude I0 during each half cycle, where I0 = constant load current.

is t   I 0   Cn sin( nt   n 
n 1, 2,3
where Cn  a 2 n  b 2 n
a 
and  n  tan 1  n 
 bn 
Now
2 

1  
I 0 .d (t )   I 0 .d (t )  0


2  
 

I0 
an 
2  

1  
I 0 cos ntd (t )   I o cos nt.d (t )


 
 

4Io
sin n ............. for n  1,3,5.....
n
 0............................. for n  2,4,6......

bn 
2  

1  
I
sin
n

t
.
d
(

t
)

I o. sin nt.d (t )


o

 
n 

4I o
cos n ............. for n  1,3,5.....
n
 0............................. for n  2,4,6......

1/ 2
2
2
 4 I
  4I o
 
o
C n   
sin n   
cos n  
  n
 
 n
4I
 o
n
 n  tan 1  tan n   n
= displacement angle of nth harmonic current.
4I o
sin( nt  n )
 is (t )   n

n 1,3,5
The rms value of nth harmonic input current, from the above equation is
i sn 
4I o
2 2 Io

n
2.n
2 2 .I o

rms fundamental circuit, i s1 
1/ 2
 I 2 0 . 
rms value of total input current Is  

  
 I0
also 1  
-ve sign shows that fundamental current lags the source voltage.
Displacement factor (DF)
DF  cos1  cos( )  cos
Current distortion factor (CDF)
CDF 
I S1 2 2 .I o 1 2 2


 0.90032
IS

Io

Harmonic factor (HF) or Total harmonic distortion (THD)
 1

HF or THD  
 1
2
 CDF

powerfactor  CDF  DF 
1/ 2
   2 
 
  1
 2 2 

2 2
cos

 V
voltage ripple factor   0 r
 V0
2


  1


1/ 2
1/ 2
 0.48343 or 48.343%
Active Power
Power input Pi  Vs .I s1. cos 1
 Vs .
2 2 .I o
2V .I
cos   m o cos 


 V0 .I o
Reactive Power
Reactive Power Qi  Vs .I s1.sin 
 Vs .
Qi 
2 2 .I o
2V .I
sin   m o sin 


Vo
I 0 sin   Vo .I o tan 
cos 
PERFORMANCE PARAMETERS OF SEMI CONVERTER
For this semi converter, the variation of input current is is continuous from α to Π,
(Π+α) to 2Π and so on. But is is not constant & assumed to be ripple free with its amplitude
Io.
Io 
2

1 
I
.
d
(

t
)

I 0 .d (t )  0


0

2 
 

an 
2

1 
I
cos
n

td
(

t
)

I o cos nt.d (t )


0

 
 

2I o
sin n ............. for n  1,3,5.....
n
 0............................. for n  2,4,6......

bn 
2

1 
I
sin
n

t
.
d
(

t
)

I o. sin nt.d (t )


o

 
n 

2I o
(1  cos n )............. for n  1,3,5.....
n
 0............................. for n  2,4,6......

2
2
 2 I
  2Io
 
o
C n   
sin n   
(1  cos n )  
  n
 
 n
2 2
(1  cos n )1 / 2
n
4I o
n

cos
n
2

1/ 2
 an 
sin n 
  tan 1  

 1  cos n 
 bn 
 n  tan 1 
n
2
-
4I o
n
n 

cos
. sin  nt 

2
2 
n 1,3,5 n


 i s (t )  
i sn 
4I o
n 2 2 I o
n
cos

cos
2
n
2
2.n
rms fundamental circuit, i s1 
2 2 .I o

cos

2
 (   ) 
rms value of total input current Is   I o 2
 

1/ 2
1  
 (   ) 
 I0 
  

2
displacement factor (DF)


DF  cos 1  cos(  )  cos
2
2
Current distortion factor (CDF)

2 2 cos
I
2 2 .I o


2
CDF  S1 
cos

IS

2 Io   
 (   )
1/ 2
1/ 2
 1

HF or THD  
 1
2
 CDF



  (   ) 

 1
2
 8 cos

2


1/ 2
  (   )


 1
 4(1  cos  ) 
2 2 cos
powerfactor  CDF  DF 

2  cos 
2
 (   )
1/ 2
1/ 2


2
 2

  (   ) 
1/ 2



2
cos

2   (   ) 
2
.(1  cos  )
Active Power
Power input Pi  Vs .I s1. cos 1
 Vs .

2 2 .I o

 V .I . 2
cos . cos  s o
(1  cos  )

2
2

Vm
(1  cos  )  V0 .I o

Reactive Power
Reactive Power Qi  Vs .I s1. sin 1
 Vs .

Also,
2 2 .I o

 V .I
cos . sin  m o sin 

2
2

1
Re active power required in 1 phase full converter for the same Io 
2
Vo 
Qi 
Vm
V
Vo
(1  cos  ) or m 


(1  cos  )
Vo .I o

sin   Vo .I o tan
1  cos 
2
EFFECT OF SOURCE INDUCTANCE
In the analysis presented above, it has been assumed that the source has no
impedance. In actual practice the converter is invariably connected to ac mains through a
transformer so that the voltage input to the converter can be adjusted to the desired value. The
leakage impedance of the transformer (i.e source) should be taken into account for exact
analysis.
The source impedance is partly resistive and partly inductive. The effect of source
resistance is to reduce the input voltage
by an amount equal to voltage drop across
resistance. Since source resistance is small this resistance voltage drop can usually be
neglected.
However the effect of source inductance must be taken into account. An important
property of inductance is that current through an inductance cannot change instantaneously.
This affects converter operation. In a converter, the current is transferred from one thyristor
to another frequently. Because of source inductance the current in the outgoing thyristor
cannot change from full value to zero instantaneously. moreover, the current through the
incoming thyristor cannot increase from zero to full value instantaneously. Therefore after
the triggering gate pulse is applied to a thyristor, the current of the outgoing thyristor
decreases from full value to zero over a time ωt = µ. During this time interval the current
through incoming thyristor rises from zero to full value. During this period µ known as
commutating period, both the outgoing and incoming thyristors are conducting. µ is
also known as overlap angle. The overlapping of currents causes a reduction in output
voltage.
Single phase fully controlled bridge converter.
(a) Single phase fully controlled bridge converter circuit diagram
Figure shows a single phase fully controlled bridge convereter with source inductance Ls.
The load is assumed to be highly inductive so that load current can be assumed to be constant
and equal to I0. Let i1 and i2 be the currents through Th1, Th2 combination and Th3, Th4
combination respectively. During overlap period µ one of these currents decays to zero and
the other builds up from zero to full value. Fig.(b) shows the waveforms.
(b) Voltage and current waveforms
During the overlap period, the two branches in fig(a) are in parallel. Therefore,
v1  Ls
di1
di
 v2  Ls 2
dt
dt
 di di 
or v1  v2  Ls  1  2 
 dt dt 
let v1  Vm sin t
then
v 2  Vm sin t
 di di 
therefore Ls  1  2   v1  v2  2Vm sin t
 dt dt 
sin ce i1  i2  I 0  cons tan t
di1 di2

0
dt dt
di V sin t
i.e 1  m
dt
Ls
Th1, Th2 pair starts conducting at ωt = α and current through them builds up from zero at ωt =
α to I0 at ωt = α+µ. Therefore from t = α/ω to t = (α+µ)/ω the current builds up from zero to
I0.
Therefore
I0
( +µ)/ 
0
 /
 di1 

I0 
or
Vm sin t
dt
Ls
Vm
cos   cos(   )
Ls
The output voltage is zero from ωt = α to ωt = (α+µ). Therefore to find average output
voltage , the effective period of conduction is (α+µ) to (Π+α). Therefore
V0 
Vm  +
 sin t d(t)
  +µ
V0 
Vm
cos(   )  cos(  )

V0 
or
Vm
cos(   )  cos  

From above equation
cos(   )  
Ls
Vm
I 0  cos 
Substituting the value of cos(α+µ) in the above equation we get,
V0 
2Vm
L
cos   s I 0


2Vm
cos 
If the effect of source inductance is neglected , the average output voltage is 
Ls
I0
Thus the source inductance causes a voltage drop equal to 
GENERATION OF GATE TIMING PULSES FOR SINGLE PHASE CONTROLLED
RECTIFIERS.
A gate trigger circuit for thyristors in phase controlled rectifiers should possess the
following:
(i)
A circuit for the detection of zero crossing of the input voltage
(ii)
Generation of trigger pulses of required wave shape.
(iii)
DC power supply for pulse amplifier
(iv)
Gate trigger circuit isolation from the line potential by means of pulse
transformers or opto couplers.
A general block diagram for gate trigger circuit for single phase converter is shown in
figure(a). The gating circuit consists of synchronising transformer, diode rectifier, zero
crossing detector, firing angle delay block, pulse amplifier, gate pulse isolation transformer
and power circuit for the converter.
Fig.(a) Block diagram of a thyristor gate circuit
Synchronizing centre tapped transformer steps down the supply voltage suitable
for zero crossing detector and for delivering dc supply Vcc to gate trigger circuit.
the zer crossing detector converts ac synchronizing input voltage into ramp
voltage and synchronizes this ramp voltage with zero crossing of the ac supply
voltage as shown in figure (b)
(b) Waveforms for the gate circuit
In the firing angle delay block the constant amplitude ramp voltage is
compared with control voltage EC. when rising ramp voltage equals control voltage EC, a
pulse signal of controlled duration is generated as shown in fig(b). these signals are indicated
as vi for thyristors 1&2, and vj for thyristors 3&4 for the power circuit of fig(a). If EC is
lowered, firing angle decreses and in case EC is raised, firing angle increases. This shows that
firing delay angle is directly propotional to the control signal voltage. The pulse output from
the firing delay angle block are next fed to a pulse amplifier circuit. the amplified pulses are
then used for triggering thyristors 1,2,3 and 4 through gate pulse isolation transformers as
shown in figure.
THREE-PHASE CONTROLLED CONVERTERS
The converter operating from a single-phase supply produces a relatively high proportion
of a.c ripple-voltage at its d.c terminals. This ripple is generally undesirable because of its
heat producing effect. Therefore, a large outlay of smoothing reactor is necessary to
smoothen the output voltage as well as to reduce the possibility of discontinuous operation.
The need for smoothing can be minimised by increasing the number of pulses. A three phase
a.c supply with a suitable transformer connection permits an increase in the pulse number.
When the number of pulses of the converter is increased, the number of segments that
fabricate the output voltage also increases and consequently the ripple content
decreases. Higher the pulse number, smoother is the output voltage.
Three-phase rectifier circuits are used for large power applications. Generation of the
three-phase a.c. Power is now universal and in some countries, only generation frequencies
may be different. Now-a-days, 11kV, 33 kV, 66kV three-phase a.c. Supply is available to the
industries. These voltages are suitably stepped down using transformers. These transformers
are generally delta-connected on primary side and star-connected on the secondary side.
Three-phase controlled converter circuits can be studied under following categories :
(1) Three-pulse converters
(2) Six-pulse converters
(3) Twelve-pulse converters
THREE-PULSE CONVERTERS (M3 CONNECTION)
Three pulse converters are also known as the three-phase
half-wave controlled
rectifier . The simplest type of phase-controlled converter operating from a three-phase
supply is the three-pulse midpoint converter.
Three-Phase Half-Wave Controlled Rectifier with Resistive Load
Figure (I) shows the power-diagram of a three-phase half-wave controlled rectifier with
resistive load. This configuration is called as the mid-point configuration because all the
phase emfs can have a common terminal which may considered as the neutral point or the
mid-point. As shown in figure, the primary is connected in a delta fashion and secondary in
star. The load is connected to the neutral point. For the analysis of the circuit, the leakage
inductance and on state SCR drops are assumed to be zero. The wave forms are shown in figure
(II).
3 phase half wave thyristor converter feeding R load
(a) line to neutral source voltage, load voltage waveform for (b) 0 < α < 30° and (c) α > 30°
If firing angle is zero degree, SCR T1 would begin conducting from ωt = 30º to 150º , T2 from ωt =
150º to 270º and T3 from ωt = 270º to 390º and so on. In other words firing angle for this controlled
converter would be measured from ωt = 30º for T1, from ωt = 150º for T2 and from ωt = 270º for T3
as in figure II(a) . For zero degree firing angle delay, thyristor behaves as a diode. The operation of
this converter is now described for α < 30º and for α > 30º.
Firing angle <30º
The output voltage waveform v0 for firing angle <30º (say 15º) is shown in fig.II(b) where T1
conducts from ωt = 30º +α to ωt = 150º +α, T2 from ωt = 150º +α to ωt = 270º +α and so on. Each
SCR conducts for 120º. The waveform of load current i0 would be identical with voltage wave form
v0.
Average value output vol tage V0 

3
2

5
6
 Vm p sin td (t )


6
3 3
3V
Vm p. cos   ml . cos 
2
2
Where Vmp = maximum value of phase (line to neutral) voltage
Vml
α
= maximum value of line voltage = 3.Vmp
= firing angle delay
average load current I 0 
V0 3Vml

. cos 
R 2R
Rms value of output , or load voltage is


3V 2 mp  2
3

cos 2 

4  3
2

1 3 3

or Vor  Vmp  
cos 2 
 2 8

1/ 2
1

3
 3Vmp  
cos 2 
 6 8

1/ 2
1

3
 Vm l  
cos 2 
 6 8

1/ 2
Rms load current

V
V 1
3
Ior  or  m l  
cos 2 
R
R  6 8

Firing angle >30°
When firing angle is more than 30°, T1 would conduct from 30°+α to 180°, T2 from 150°+α
to 300° and so on in fig.II(b). For R load, when phase voltage va reaches zero at ωt=180°,
current io =0, T1 is therefore turned off. Thus T1 would conduct from 30°+α to 180°. Same is
true for other SCRs. This shows that each SCR, for firing angle >30º, conducts for (150°-α)
only. This also implies that for R load, maximum possible value of firing angle is 150º.
Waveform of io agrees with vo waveform, fig.II(c).
Average value of load voltage:
V0 
3 
 Vm p sin td (t )
2 


3Vm p
2
6
1  cos(  30)
1/ 2


 3  2

2
RMS value of output vol tage Vor  
V m p sin t.d (t )

2 



6
1/ 2
3.Vmp  5

 1
Vor 
 6     2 sin( 2   / 3)
2  



V  5
 1
 ml 
    sin( 2   / 3)
2   6
 2

1/ 2
Three Phase M-3 converter with R-L Load
In this the load is R-L type. The load inductance L is large so that load current is continuous
and constant at Io as shown in figure below. For firing angle < 30°, Vo and Vor are given as:
Average value output vol tage V0 
3Vml
. cos 
2
1/ 2
1

3
RMS value of load voltage Vor  Vm l  
cos 2 
 6 8

For the firing angle range of 30º < α < 90° and 90º < α < 180°, this converter behaves
differently as described below.
30º < α < 90° : for firing angle in this range, let the firing angle be say 45º for which the
various waveforms are drawn. In this T1 conducts from 30º+α to 150º+α, T2 from 150º+α to
270º+α, T3 from 270º+α to 390º+α and so on. Thus each SCR conducts for 120º.
At ωt = Π, phase voltage va is zero, but iT1(or ia) is not zero because of RL load. Therefore T1
would continue conducting beyond ωt = Π, as such vo = va goes negative beyond ωt = Π.
When T2 is turned on at ωt = 150º+Π, load current shifts from T1 to T2 and a voltage va-vb
=( Vm sin(150+α)-Vm sin(30+α) ) appears as reverse bias across T1 to aid its communication.
SCR T2 conducts from (150+α) to (270+α) and so on. The waveform for iT1 or ia, iT2 or ib and
iT3 or ic are as shown in figure.
The wave form vT1 for voltage across T1, on the assumption of firing angle 45º can be drawn
as under:
When T1 is on, vT1 = va-va=0 from ωt = 75º to 195º
When T2 is on, vT1 = va-vb from ωt = 195º to 315º
When T3 is on, vT1 = va-vc from ωt = 315º to 435º and so on
When T2 is turned on at ωt = 195º, vT1 = va-vb = -Vmp sin 15º- Vmp sin 75º = -1.225 Vmp:
at ωt = 210º, vT1 = -1.5Vmp:
at t  240o vT 1  3.Vm p; at t  270o , vT 1  - 1.5Vm p;
at t  300o , vT 1  - Vm p sin 60  0  0.866Vm p;
at t  315o , vT 1  - Vm p sin 45o  Vm p sin 15o  0.448Vm p
Also at ωt = 315º, T2 gets turned off whereas T3 is turned on.
vT1 = va-vc = -Vmp sin 45º+ Vmp sin 75º = -1.673 Vmp:
this shows that at at ωt = 315º, vT1 at once changes from -0.448 Vmp to -1.673 Vmp: as shown
in figure.
Three phase M-3 converter waveforms for 30° < α < 90° for ripple free load current
at t  330o vT 1  - Vm p sin 30  Vm p  - 1.5Vm p;
at t  360o , vT 1  0 - 0.866Vm p  0.866Vm p;
at t  390o , vT 1  0.5Vm p  0.5Vm p  0
at t  420o , vT 1  0.866Vm p  0  0.866Vm p;
at t  435o , vT 1  Vm p sin 75o  Vm p sin 15o  1.225Vm p
and also v T 1  va  va  0 and so on.
90º<α<180º:
For firing angle in this range, let α be say 165º. The various waveforms are shown in figure
below.
Three phase M-3 converter waveforms for 90° < α < 180° for ripple free load current
As the output voltage waveform vo is below the reference line, average value of vo
must be negative. It is also evident from Vo = (3Vml/2Π)cosα that when firing angle α is more
than 90º, Vo is negative. For α>90º, three phase 3 pulse converter operates as line
commutated inverter which is possible only if the load circuit has a dc voltage source of
reverse polarity, as in a single phase full converter.
Three phase Full Converter
The figure shows a 6 pulse bridge converter.this converter is most widely used in industrial
applications upto the 120kW level, where two quadrant operation is required.
3 phase fully controlled Bridge convereter
The load is fed via 3 phase half wave connection to one of the 3 supply lines, no
neutral being required. Hence transformer connection is optional. However, for isolation of
output from the supply source, or for higher output requirements, the transformer is to
be connected. If transformer is used, then one winding is connected in delta because the delta
connection gives the circulating path for third harmonic current. Therefore, third harmonic
current doesnot appear in line which is an advantage.
This circuit consists of two groups of SCRs, positive group and negative group. Here,
SCRs T1, T3, T5 forms positive group, whereas SCRs T4,T6,T2 forms a negative group. The
positive group SCRs are turned on when the supply voltages are positive and negative group
SCRs are turned on when the supply voltages are negative. In order to start the circuit
functioning, two thyristors must be fired at the same time in order to commence current flow,
one of the upper arm and one of the lower arm.
For describing the operation of the circuit, the following things to be remembered.
(i)
Each device should be triggered at a desired firing angle α
(ii)
Each SCR can conduct for 120º
(iii)
SCR must be triggered in the sequence T1,T2,T3,T4,T5,T6
(iv)
The phase shift between the triggering of the two adjacent SCR is 60º
(v)
At any instant 2 SCRs can conduct and there are such 6 pairs. The 6 pairs are
(T6,T1), (T1,T2), (T2,T3), (T3,T4), (T4,T5), (T5,T6).
(vi)
Each SCR conducts in two pairs and each pair conducts for 60º
(vii)
The incoming SCR commutates the outgoing SCR,i.e SCR T1 commutates SCR
T5, SCR T2 commutates SCR T6 and so on.
(viii) When the two SCRs are conducting, i.e one from +ve (upper) group and one from
–ve (lower) group, the corresponding line voltage is applied across the load.
(ix)
When the upper SCR of a half bridge conducts, the current of that phase is +ve
whereas when the lower SCR conducts, the current is –ve.
Three phase Full Converter with Resistive load
Three phase fully controlled bridge rectifier with resistive load is shown in figure.
(a) 3 phase fully controlled bridge rectifier with resistive load
For six pulse operation, each SCR has to be fired twice in its conduction cycle, that is
firing intervals should be 60º. The ouput voltage waveforms for different values of α are
shown in figure (b)
(b) Voltage waveforms for various firing angles
(c)
The following points can be noted:
(1) The output voltage waveform for any value of α is a 6 pulse wave with a ripple
frequency of 300Hz.
(2) Continuous conduction mode (0 ≤ α ≤ Π/3) when the phasor AB is allowed to conduct
at α between 0 to Π/3, it continuous to conduct by 60º when the phasor AC is fired.
The conduction is shifted from SCR T6 to T2. T6 is commutated off by the reverse
voltage of phase C and B across it. The phasor AC conduct after another 60º after
which it is replaced by phasor BC when phase B voltage assumes greater value than C
or A. hence load current is continuous for α between 0 to Π/3.
(3) Discontinuous conduction mode : (Π/3 ≤ α ≤ 2Π/3)
When Π/3 ≤ α ≤ 2Π/3, the phasor AB conducts upto an angle Π after which both the
thyristors T1 & T6 are commutated off because phase B becomes +ve w.r.to phase C
and after 60º, when T2 & T1 are fired, phase AC conducts also upto angle Π, hence
load current remains zero from angle Π to the next firing pulse and becomes
discontinuous, therefore the fully controlled bridge circuit produces a ripple frequency
of 6 times the supply frequency at all trigger angles.
(4) For α =120º, the ouput voltage is zero and hence αmax = 120(2Π/3)
(a) Continuous conduction mode: (α <60º). The general equation for the average load
voltage is given by
Edc 
1 2
 Edc (t ).d (t )
2 0

Edc

1 2
 6
 E AB (t ).d (t )
2 
6

Where the line to line Voltage EAB is given by
EAB  3Em sin( t   / 6)
 Edc 


2
3


6

3Em sin( t   / 6).d (t ) 

3 3 Em
cos 

Average load current I d 
3 3 Em
cos 
.R
3 3Em

2

3
 sin( t ).d (t )


3
(b) Discontinuous conduction mode (α˃60°)
5
1 6
Edc  6 

2 
6

3Em sin( t   / 6)d (t ) 

3 3Em 
 sin( t ).d (t
 
3

3 3Em
(1  cos(   / 3))

or  max, Edc  0,

3 3 Em
(1  cos(   / 3))  0

hence  max  1200
Average load current I d 
3 3 Em
(1  cos(   / 3))
.R
Three phase Full Converter with R-L load
The power diagram of the 3 phase fully controlled converter with RL load is shown in
figure.
3 phase fully controlled bridge rectifier with R-L load
The output voltage waveforms with firing angles are shown in figure below. The load
inductance is assumed to be very large so as to produce a constant load current.
3 phase fully controlled rectifier - Output waveforms with R-L load
Following points can be noted from the waveforms.
(i)
Waveforms are similar with R load for α = 0°, 30° and 60°.
(ii)
For α˃60°, the waveforms are different. The voltage goes –ve due to the
inductive nature of the load.the previous thyristor pair continuous to conduct
till the next SCR is triggered.
(iii)
For α =90°, the area under the +ve and the –ve cycle are equal and the average
voltage is zero.
(iv)
For α˂90°, the average output voltage is +ve and for α˃90°, the average ouput
voltage is negative.
(v)
The maximum value of α is 180°
(vi)
The output is always a six pulse, i.e ripple frequency is 300Hz irrespective of
the value of α
Expression for Average and RMS output voltage
Average output vol tage E dc  6 

E dc 
1 90
 ERY (t ).d (t )
2 30
3 90
3 120
3
E
sin(

t

30
).
d

t


 3Em sin( t ).dt
m
 30
 60
3 3 Em
cos  for 0    180o

Em is the peak value of the line to neutral voltage.
As the firing angle changes from 0 to 900, the voltage also changes from
maximum to zero and the converter is said to be in rectification mode.
For the angles in the range 90° to 180°, the voltage varies from 0 to –ve
maximum and the converter is in the inversion mode.
RMS output voltage
1/ 2
 1 2 2

E rms  
E dc (t ).d (t )

 2 0

1/ 2


1 90 2
 6 
E AB (t ).d (t )

 2 30

1/ 2
 3 90

9 E 2 m 120 2
2

( 3Em sin( t  30)) .dt 
sin (t ).dt 


 60
  30

1/ 2
 9 E 2 m 120


1  cos( 2t ).dt 

 2 60

1/ 2

3E  2
3
 m 
cos 2 
2 3 

for 0    180o