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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

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