This document discusses multiphase star rectifiers and three-phase bridge rectifiers. It provides the following key points: 1) Multiphase star rectifiers use multiple phases (q phases) with each diode conducting for 2π/q periods to rectify the input. They can be considered as multiple single-phase half-wave rectifiers. 2) Three-phase bridge rectifiers provide full-wave rectification with six pulses on the output voltage from the conduction sequence of diodes. They have higher output voltage and lower ripple compared to star rectifiers. 3) Equations are given for the output voltage and current characteristics of both rectifier types with resistive and RL loads.

1. Multiphase star rectifier
• For larger (>15kW) output power.
• Harmonics and fundamental component.
• Filter size decreases with the increase of
harmonics.
• Q phase filter has fundamental component
of qf frequency.
• Also known as star rectifier.

2. • May be considered as q
single-phase half-wave
rectifier.
• K-th diode conducts during the
period when the voltage of k-th
phase is greater than that of
other phases.
• The conduction period of each
diode is 2π/q.
• Primary must be delta
connected to compensate the
dc component flowing through
secondary windings.

3. (1)
.....
sin
sin
)
(
cos
/
2
2
/
0
/
0
q
q
V
t
q
V
t
d
t
V
q
V m
q
m
q
m
dc











 
(2)
.....
2
sin
2
1
2
4
2
sin
2
)
(
cos
/
2
2
2
/
1
/
0
2
/
1
/
0
2






































  q
q
q
V
t
t
q
V
t
d
t
V
q
V m
q
m
q
m
rms











Assuming a cosine wave from π/q to 2π/q, the average o/p voltage is.
Output rms voltage is,
If the load is purely resistive then peak current through the diode is
Im = Vm/R and the rms current through a diode is,
2
/
1
/
0
2
2
)
(
cos
2
2






 
q
m
s t
d
t
I
I




(Integrating over the whole period, 2π)
R
V
q
q
R
V
q
q
I
I s
m
m
s 
































2
/
1
2
/
1
2
sin
2
1
2
1
2
sin
2
1
2
1 





Vs is the rms voltage of a transformer secondary

4. Three phase bridge rectifiers
• Full-wave rectifier.
• Gives six-pulse ripples on
the o/p voltage.
• Conduction sequence of
diodes are,
12,23,34,45,56,61.
• The pair of diodes which
are connected between
that pair of supply lines
having the highest line-to-
line instantaneous
voltage will conduct.
• Line to line voltage,
p
ij V
V 3


5. • Average o/p voltage,


6
/
0
)
(
cos
3
6
/
2
2 



t
d
t
V
V m
dc
m
m V
V 654
.
1
3
3



(Vm is the peak phase voltage)
Rms o/p voltage, 2
/
1
6
/
0
2
/
1
6
/
0
2
2
4
2
sin
2
3
.
6
)
(
cos
3
6
/
2
2






















 

 





t
t
V
t
d
t
V
V m
m
rms
m
m
m
m V
V
V
V 6554
.
1
4
3
9
2
3
2
3
.
4
18
12
18
6
2
sin
4
1
2
6
/
18
2
/
1
2
/
1






































For purely resistive load, peak diode current, R
V
I m
m /
3

Rms diode current,
2
/
1
6
/
0
2
/
1
6
/
0
2
2
2
sin
4
1
2
2
)
(
cos
2
2
.
2






















 








t
t
t
d
t
I
I m
r
m
m I
I 5518
.
0
6
2
sin
2
1
6
1
2
/
1



















6. Rms value of transformer secondary current,
2
/
1
2
/
1
6
/
0
2
2
6
2
sin
4
1
6
.
2
4
)
(
cos
2
4
.
2 



















 







m
m
s I
t
d
t
I
I
m
m I
I 7804
.
0
6
2
sin
2
1
6
2
2
/
1


















Where Im is the peak secondary line current = peak diode current,
R
V
I m
m /
3


7. 3-phase bridge rectifier with RL
load
t
V
E
Ri
dt
di
L ab
L 
sin
2



3
2
3
for
sin
2



 

 t
t
V
v ab
ab
81)
-
(3
...
)
sin(
2 )
(
1
R
E
e
A
t
Z
V
i
t
L
R
ab
L 






Where load impedance   2
1
2
2
)
( L
R
Z 


and load impedance angle .
tan 





 
R
L


The constant A1 can be found from the initial condition: at .
,
3 1
I
i
t L 
 

  
  
  












3
1
1
3
1
1
3
1
1
3
sin
2
3
sin
2
3
sin
2
L
R
ab
L
R
ab
L
R
ab
e
Z
V
R
E
I
A
e
A
Z
V
R
E
I
R
E
e
A
Z
V
I







































8.       
 
 
82)
-
(3
.
.
.
3
sin
2
sin
2
3
sin
2
sin
2
3
1
3
1
R
E
e
Z
V
R
E
I
t
Z
V
i
R
E
e
e
Z
V
R
E
I
t
Z
V
i
t
L
R
ab
ab
L
L
R
t
L
R
ab
ab
L


























































.
)
3
/
(
)
3
/
2
( 1
I
t
i
t
i L
L 


 



Under steady-state condition,
 
 
 
83)
-
(3
.
.
.
0
I
for
)
1
(
3
sin
sin
2
)
1
(
3
sin
sin
2
1
3
sin
sin
2
)
1
(
1
3
3
1
3
2
3
3
2
3
1
3
2
3
3
2
3
3
2
3
1

















































































































































R
E
e
e
t
Z
V
I
R
E
e
e
t
Z
V
I
e
R
E
e
t
Z
V
e
I
L
R
L
R
ab
L
R
L
R
ab
L
R
L
R
ab
L
R





































9. Which after substituting in eq (3-82) and then simplifying, gives
 
 
84)
-
(3
...
0
and
3
/
2
/3
for
)
1
(
3
sin
3
2
sin
sin
2 3
3























































L
L
R
L
R
ab
L
i
t
R
E
e
e
t
Z
V
i








 



Rms diode current can be found from above equation as,
2
/
1
3
2
3
/
2
)
(
2
2






 




t
d
i
I L
r
Rms o/p current is the combination of rms current of each diode.
r
rms I
I 3


10. Effects of source and load
inductance
.
angle
n
commutatio
called
is
which
time
some
for
conducts
diodes
both
and
equal
are
D3
and
D1
of
voltage
anode
is
Result
.
is
to
due
voltage
o/p
time
same
At the
.
becomes
o/p
and
of
L1
across
voltage
induced
and
resulting
decreases,
current
The
.
D
through
flows
still
and
equal
are
and
at
2
1
1
1
1


L
bc
L
bc
L
ac
L
L
d
dc
bc
ac
v
v
v
v
v
v
v
v
i
I
v
v




