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                                                            Power electronic control in electrical systems 275

                      equation (7.11), beginning at the time when the thyristor is turned on, at f   s 1 /2,
                                                                                      1
                      and with an initial condition that the current is zero
                                                       1  Z  t
                                              I R (ot) ˆ      V(ot)dot                   (7:29)
                                                      L R   s 1
                                                          f 1
                                                             2
                      If the voltage is assumed to be a harmonic series then the current will have a similar
                      representation
                                                 1           jhot        t
                                                X    1     e     e   jhot
                                        I R (ot) ˆ      V                                (7:30)
                                                   hoL R        j2         s 1
                                                 1                      f 1    2
                      This equation describes the current until the thyristor turns off. Just after the first
                      zero crossing, at f ‡ s 1 /2 ! I R (ot) ˆ 0
                                      1
                                                  1
                                                 X    2V          hs 1
                                              0 ˆ         e  jhf 1  sin                  (7:31)
                                                     hoL R         2
                                                   1
                      Similarly
                                                  1
                                                 X    2V          hs 2
                                              0 ˆ         e  jhf 2  sin                  (7:32)
                                                     hoL R         2
                                                   1
                      The constrained equations (7.31) and (7.32) are solved by iteration. Each is depen-
                      dent on two angles, s 1 and f for equation (7.30) and s 2 and f for equation (7.31).
                                               1
                                                                            2
                      The angles s 1 and s 2 may be used to determine the switching vector in equation
                      (7.16).
                      7.4.3   Three-phase TCRs

                      A three-phase TCR normally consists of three delta connected, single-phase TCRs in
                      order to cancel out the 3rd, 9th and 15th harmonic currents. Banks of capacitors and
                      TSCs produce no harmonic distortion and there is no incentive for them to be
                      connected in delta. The admittance matrix of equation (7.27) can be used as the basic
                      building block for assembling three-phase TCR models. Linear transformations can
                      be used for such a purpose.
                        The combined harmonic equivalent of three single-phase TCRs is
                                        0    1    0                 10    1
                                          I R,1     H R,1  0     0      V 1
                                        @  I R,2  A  ˆ  @  0  H R,2  0  A@  V 2  A       (7:33)
                                          I R,3      0     0    H R,3   V 3
                      In a power invariant, delta connected circuit the relationships between the uncon-
                      nected and connected states are

                                         0   1        0             10    1
                                           I A           1   0    1     I 1
                                                 € 30
                                         @  I B  A  ˆ p   @   1  1  0  A@  I 2  A     (7:34)
                                                    3
                                           I C           0   1    1     I 3
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