180 degree conduction
DESCRIPTION
inverterTRANSCRIPT
Three-Phase Inverter (continued)
Three single-phase full bridge inverters
12 transistors, 12 diodes, 3 transformers
Could it be simpler?
Mode 1 Operation
1
1
1
3
2 22
3
2 32
3
eq
s s
eq
san cn
sbn
R RR R
V Vi
R R
Vi Rv v
Vv i R
03
t
Q1, Q5, Q6 conduct
Mode 2 Operation
2
2
2
3
2 22
3
2
3
2 3
eq
s s
eq
san
sbn cn
R RR R
V Vi
R R
Vv i R
Vi Rv v
2
3 3t
Q1, Q2, Q6 conduct
Mode 3 Operation
2
3t
3
3
3
3
2 22
3
22
3
eq
s s
eq
an bn
scn
R RR R
V Vi
R R
iv v
Vv i R
Q1, Q2, Q3 conduct
Fourier Series for Line-to-Line Voltages
1
5 5
6 6
5
6 6
1,3,5,...
( cos( ) sin( ))2
1( ) ( )
4sin( )sin( )
2 34sin sin ( )
3 6
oab n n
n
n s s
sn
sab
n
av a n t b n t
b V d t V d t
V n nb
nV n
v n tn
For the other Line-to-Line Voltages
1,3,5,...
1,3,5,...
4sin sin ( )
3 2
4 7sin sin ( )
3 6
sbc
n
sca
n
V nv n t
n
V nv n t
n
rms value of the nth Component
1
4sin
321
4 sin 600.7797
2
sLn
sL s
V nV
nn
VV V
Fundamental Component
Phase Voltages (Y-connected load)
1,3,5,..
1,3,5,..
1,3,5,..
4sin( )sin( )
33
4 2sin( )sin ( )
3 33
4 4sin( )sin ( )
3 33
saN
n
sbN
n
scN
n
V nv n t
n
V nv n t
n
V nv n t
n
DC Supply Current
IL=√3Io is the rms load line current
Vo1= fundamental rms output line voltage
Io is the rms load phase current
Θ1 = the load impedance angle at the fundamental frequency
11
11
( ) ( ) ( ) ( ) ( ) ( )
....
3 cos( )
3 cos( )
s s ab a bc b ca c
os o
s
os L
s
v i v t i t v t i t v t i t
VI I
V
VI I
V