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)
 
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