Direct Power and Torque Control of AC/DC/AC Converter-Fed ...
Direct Power and Torque Control of AC/DC/AC Converter-Fed ...
Direct Power and Torque Control of AC/DC/AC Converter-Fed ...
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2. Voltage Source <strong>Converter</strong>s – VSC<br />
I<br />
= I S + I S I S ,<br />
(2.59)<br />
dc LA A LB B<br />
+<br />
LC<br />
C<br />
Hence, <strong>AC</strong> side <strong>of</strong> the VSR voltage equations in the three phase system can be<br />
expressed as:<br />
dI<br />
L<br />
dt<br />
LA<br />
dI<br />
L<br />
dt<br />
LB<br />
C<br />
⎛ 1<br />
+ RILA<br />
= U<br />
LA<br />
−Udc⎜<br />
S<br />
A<br />
− ∑ S<br />
⎝ 3 k = A<br />
C<br />
⎛ 1<br />
+ RI<br />
LB<br />
= U<br />
LB<br />
−U<br />
dc⎜<br />
SB<br />
− ∑ S<br />
⎝ 3 k = A<br />
k<br />
⎞<br />
⎟ = U<br />
⎠<br />
k<br />
LA<br />
⎞<br />
⎟ = U<br />
⎠<br />
LB<br />
−U<br />
dc<br />
−U<br />
⎛<br />
⎜ S<br />
⎝<br />
dc<br />
A<br />
⎛<br />
⎜ S<br />
⎝<br />
B<br />
1<br />
⎞<br />
− ( S<br />
A<br />
+ SB<br />
+ SC<br />
)⎟<br />
3<br />
⎠<br />
(2.60)<br />
1<br />
⎞<br />
− ( S<br />
A<br />
+ SB<br />
+ SC<br />
)⎟<br />
3<br />
⎠<br />
(2.61)<br />
dI<br />
L<br />
dt<br />
LC<br />
C<br />
⎛ 1<br />
+ RI<br />
LC<br />
= U<br />
LC<br />
−U<br />
dc⎜<br />
SC<br />
− ∑ S<br />
⎝ 3 k = A<br />
Where the line voltages are expressed by:<br />
U<br />
LA<br />
= U<br />
Lm<br />
sin<br />
k<br />
⎞<br />
⎟ = U<br />
⎠<br />
LC<br />
−U<br />
dc<br />
⎛<br />
⎜ S<br />
⎝<br />
C<br />
1<br />
⎞<br />
− ( S<br />
A<br />
+ SB<br />
+ SC<br />
)⎟<br />
3<br />
⎠<br />
(2.62)<br />
( ω t) , U = U sin( ω t − 2 π / 3) , U = U sin( ω t + 2π<br />
/ 3)<br />
L<br />
LB<br />
Also <strong>DC</strong>-link side equation <strong>of</strong> the VSR can be modeled by:<br />
dU<br />
C<br />
dt<br />
dc<br />
C<br />
= ∑ I<br />
Lk<br />
k = A<br />
S<br />
k<br />
− I<br />
load<br />
= I<br />
The above expressions can be written as:<br />
d<br />
( L<br />
dt<br />
Lm<br />
LA<br />
S<br />
A<br />
+ I<br />
L<br />
LB<br />
C<br />
⎛ 1<br />
+ R )I<br />
Lk<br />
= U<br />
Lk<br />
−U<br />
dc⎜<br />
Sk<br />
− ∑ S<br />
⎝ 3 k=<br />
A<br />
k=<br />
A<br />
S<br />
k<br />
B<br />
⎞<br />
⎟<br />
⎠<br />
+ I<br />
LC<br />
S<br />
C<br />
LC<br />
− I<br />
load<br />
Lm<br />
L<br />
(2.63)<br />
(2.64)<br />
(2.65)<br />
C<br />
dU<br />
dc<br />
C = ∑ I<br />
Lk<br />
Sk<br />
− Iload<br />
(2.66)<br />
dt<br />
The equations (2.65) <strong>and</strong> (2.66) can be represented as a block diagram in Fig. 2. 15<br />
[13], [16].<br />
28