power electronics total material

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




