SENSORLESS FIELD ORIENTED CONTROL OF BRUSHLESS ...

however. Assume for example that the commanded SV is in sextant s 12 as shown in Figure 4.30.
S
S
as
The obvious choice for the active base SVs would be 1 S
and 2 ; we could arbitrarily pick 7
the null base SV. From Table 4.1, the switch states would progress as follows.
S1 S2 S7
HLL HHL HHH
Each state transition within a period requires a change of only one leg state. However, the
transition between periods requires two legs to switch states. Each time a leg changes states a
does not solve
switching loss is incurred, so this is not desirable. Selecting the null SV to be 0 S
the problem. Since inter-period transitions required only one transition, the period could be
rearranged as shown in Figure 4.36, with appropriate changes to the durations of each base SV
(dividing the previous Tx and Ty in half).
Figure 4.36 – One period in an improved switching scheme.
This solves the switch transition problem but there is another problem. Assume the commanded
SV approaches S2
and crosses over into sextant s 23 . SXS1and SYS2would become
S S2and
S S3.
As the SV crossed the boundary there would be extraneous switching
X
Y
between two periods and this could cause distortion. This problem can be solved by dividing the
null time, adding a segment at the beginning and end of each period, as shown in Figure 4.37.
Figure 4.37 – One period on a second improved switching scheme.
But distributing the null SV as shown would cause the switch states would progress as follows.
S S S S S S S
7 1 2 7 2 1 7
HHH HLLHHLHHH HHLHLLHHH
The null state next to S1 causes a double transition where previously there was none. The solution
is to use S0 for the beginning and end null states and S7 for the center null state.
S S S S S S S
0 1 2 7 2 1 0
LLLHLLHHLHHH HHLHLLLLL 194