Hardware Reference
In-Depth Information
L
AA
= L
BB
= L
CC
= L,
M
AB
= M
AC
= M
BA
= M
BC
= M
CA
= M
CB
= M and
i
A
+ i
B
+ i
C
=0.
Therefore,
[V ]=[e]+[R][i]+[L][di/dt],
(4.77)
where,
⎡
⎤
LMM
MLM
MML
⎣
⎦
[L]=
.
Moreover, the back-EMF in the armature windings of a micro PMSM used
in HDD are symmetrically sinusoidal, that is,
⎨
e
A
= K
e
ω
r
sin(ω
r
t)
e
B
= K
e
ω
r
sin
¡
¢
ω
r
t−
2
3
(4.78)
¡
¢
⎩
ω
r
t +
2
3
e
C
= K
e
ω
r
sin
Therefore,
V
AN
+ V
BN
+ V
CN
=0.
(4.79)
For the resistance circuit shown in Figure 4.81, the sum of the three currents
fl
owing into the circuits is zero,
i
1
+ i
2
+ i
3
=0,
(4.80)
which means
V
AM
+ V
BM
+ V
CM
=0.
(4.81)
Now looking back to Figure 4.81, one can easily derive the voltage equa-
tions,
⎨
V
AN
+ V
NM
+ V
MA
=0
V
BN
+ V
NM
+ V
MB
=0
V
CN
+ V
NM
+ V
MC
=0
(4.82)
⎩
By adding these three equation, one can get,
(V
AN
+ V
BN
+ V
CN
)+3V
NM
−(V
AM
+ V
BM
+ V
CM
)=0.
(4.83)
Substituting equations 4.79 and 4.81 into equation 4.83,
V
NM
=0.
(4.84)