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10_Trapezoidal_Back-Emf_PM_Synchronous_Motors_(BLDC).pdf
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详细说明：基于梯形反电动势的永磁同步电机控制技术，有原理讲解。10.Tr
idal back-Emf Pm
Motors(BLDC) 653
The magnetic axis of the rotor as well as the angle a between the north
and south poles are indicated. The permanent magnet rotor is constructed
by bonding permanent magnets on the surface of a cylindrical core of lam
inated soft iron. As usual, the rotor position is taken to be along the mag
netic axis of the rotor(see Figure 10.3). The permanent magnets on the
rotor set up a uniformly (in 0) distributed magnetic field in the air gap
given by
y
BR(,6-6R)=
T
丌
B
or
6-6R
7R(6-6R-丌/2)
一BR0r
or
22
<-6R=2
TR
3丌△
B
r
+-≤6-6R<
2
R(6-6R-3丌/2)
T
B
r for
△/2
2
f
(10.1)
where the parameter A represents the angular distance between the north
and south pole magnets on the rotor. Also, the factor re/r ensures that
conservation of fux holds in the air gap The radial magnetic field distrib
ution in the air gap due to the rotor's permanent magnet is illustrated in
Figure 10.4
Rotor magnetic field
Q
FIGURE 10.4. The radial magnetic field distribution BR(r, 0-OR)
654 10. Trapezoidal Back-Emf PM Synchronous Motors(BLDC)
In Sections 10.2 through 10.7, A will be taken to be zero to simplify the
analysis with no significant impact on the resulting mathematical model
In this case,(10.1) simplifies to
BRo-ffor-丌/2≤6-6n≤丌/2
BR(,6-6z)
(10.2)
Bro Rr for+/2≤0-0R≤3丌/2
10. 2 Stator Magnetic Field Bs
The magnetic field Bs produced stator currents is now derived. As illus
trated in Figure 10.1, the stator winding density of a bldc is uniform
over inner periphery of the stator. Specifically, stator phase 1 has a wind
ing density given by
丌
03
and
(6)
3
0
elsewhere
The total number of windings(turns or loops )making up phase 1 is then
de= m
y
Ns1(6-2丌/3) and Ns3(6)=Ns1(6-4丌/3).Tode
termine the radial magnetic field in the air gap produced by the current
in stator phase 1, Ampere's law with h=0 in the iron is applied to the
path 1-2-3-1 of Figure 10.5 to obtain by symmetry, HsI(is1, T +8)
H5s1(
N
丌/3≤6≤丌/3
Nsis
de
丌/3≤6≤2
2gH5s1(s1,)
N
2x/3≤6<4丌/3
5丌/3
N
s1
d6+
d0,4x/3<6<5m/3
T
7r/3
10. Trapezoidal Back-Emf PM Synchronous Motors(BLDC) 655
J
S1
A/ Magnetic axis of
stator phase 1
ST
FIGURE 10.5. Path 1-2-3-1 for applying Ampere's law. Note that for the path
drawn,丌/3≤θ≤2丌/3.
or
r
3≤6≤丌/3
丌
29
r for 7T/3≤b≤2丌/3
HsI(is1, 0
for2丌/3≤6≤4丌/3
Nsis 6
3
6
ffor4丌/3≤b≤5丌/3
29
2
where it was assumed that Hs1 is constant across the air gap. Then Bs1
boHsi in the air gap. A factor of rR/r is now included to ensure that Bs1
satisfies conservation of fux in the air gap. Consequently, the magnetic field
Bs1= Bs1(is1, r, 0f at any point(r, 0) in the air gap due to the current
656
0. Trapezoidal Back-Emf PM Synchronous Motors(BLDC)
isi in phase 1 is given by
N
/3≤6≤丌/3
29
0) r for T/3≤0≤2x/3
Bs1(2s1,r,0)=
polisI rR
for2/3≤6<4丌/3
uo/ rR
T
/3≤6≤5丌/3
(10.3
Similarly
Bs2(is2, r,
Bs1(s2,T,6-2m/3
Bs3(s3,r,6)=Bs1(is3,r,6-4丌/3)
a plot of the normalized radial magnetic field Bs1(is1, r, 0)/(0 9551T
due to the current in stator phase 1 for-丌/3≤≤2 T is in Figure10.6.
B
ST I I 2
36
FIGURE 10.6. Normalized magnetic field Bsi(is1, r, 0)(ga>sis ra)
This is also illustrated in Figure 10.7 which shows the direction of radial
magnetic field distribution Bs1(is1, r, 0 in the air gap at each angular
position 6. In particular, the magnetic field due to isi is radially out to the
right of the dashed vertical line and radially in to the left of the vertica
Ine
10. Trapezoidal Back-Emf PM Synchronous Motors(BLDC) 657
hOoDoo
●●●●●
Magnetic axis of
stator phase 1
FIGURE 10.7. The radial magnetic field distribution Bsi(is1, r, 0)in the air gap
due to the current isi in stator phase 1
10. 3 Stator Flux Linkage Produced by Bs
The Alux u(is1, 0) in a winding of stator phase 1 at the angle 8 where
using the flux surface shown in Figure 10. or phase 1, is now computed
3≤6≤2/3, due to the current in stat
<3 Flux
surfa
兀
o Magnetic axis
stator phase 1
7兀
U
4
3 0
丌s
FIGURE 10.8. Flux surface for a winding of stator phase 1 to calculate the flux
produced by the magnetic field of isl
658 10. Trapezoidal Back-Emf PM Synchronous Motors(BLDC)
USing the expression(10.3)for Bsl, the Alux produced by BsI in a
winding of stator phase 1 with one side at 0 and the other at 0-T is given
B
S1(2s1,7S
po/s2s1 r
-de+rsll
Plols2S1TR
S
/3
T
5丌/3
+Ts{1
YoSsi TR
3丌
d e
g丌/3Ts
10TRe1Nsis12x16/丌
3)2/5n/3
9
0+丌
PorRl1/sis
2丌3
2
T
2
TRe1Nsis1 6
25丌
2
6
direction as that chosen for positive current now in stator phase e san g
As the outward normal was used to compute the flux, -dou dt>0 mear
that this induced emf in stator phase 1 will push current in the
me
The total ux linkage Asi(is1, 0,0) in stator phase 1 due to the current
in phase 1, is then found by adding up the fluxes in each winding of phase
That is
2/3
入s1(s1,O,O)
d1(2s1,6)-d0
丌/3
21N3
2丌/3
6
7T
H
丌/3
borreINsisi
81
2
/3
borrelL
T
FOTREIT NIsi/14
3
Lsis
where(rr= e2/ 2)
L
10{12
S
The flux中21(21,6) in a winding of stator phase2 at an angle 6(丌≤0≤
47/ 3), produced by the magnetic field of stator phase 1, is computed using
10. Trapezoidal Back-Emf PM Synchronous Motors(BLDC
659
the flux surface shown in Figure10.9.withπ≤θ≤4丌/3, this flux is
computed as
sI(is1, Ts, 0).(rsde'der
lo rr
/3
de+rsll
lo Sisi TR
rsth
TS
+rsll
lo NSSl TR
de
0T1 Nsis4丌
72(-x)2x3
2丌
6
2
/3
PoTRe1/siS1
(104
Flux surface
2o
￥6
O/ Magnetic axis
stator phase 1
lL兀丌
FIGURE 10.9. Flux surface for a winding of stator phase 2 to calculate the flux
produced by the magnetic field of is1. The windings of phase 2 are shown shaded
current in stator phase 1 is ther30,0)in stator phase 2 produced by the
The total
Aux linkage入s2(2s
4m1?出31RM
/3
4丌/3
入s2(s1,O,O)=
P21(isI
s1a(丌
Misi
660 10. Trapezoidal Back-Emf PM Synchronous MotOrs(BLDC)
where(rr=e2/ 2)
M
△o12
129
The flux 31(is1, 0) in a winding of stator phase 3 at an angle 0 where
5/3<0<2T produced by the magnetic field of stator phase 1 is computed
using the flux surface shown in Figure 10.10
Flux
surfe
s
stator
7兀
FIGURE 10.10 Flux surface for a winding of stator phase 3 to calculate the fux
produced by the magnetic field of is1. The windings of phase 3 are shown shaded
Proceeding,with5丌/3≤θ≤2丌
31(2s1,θ)
Bs1(s1,Ts,6)·(rsda
0J6
4丌/3
5x/3
sll
Ao/s2S1 rR
T
2
2(2)a
+rsll
4 NsiS TRd′
5m/32g7s
(=:()
5丌/3
rl1Nsisi
5
+(6
4丌/3
slSI
(10.5)
9
The total flux linkage in stator phase 3 produced by the current in stator
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