BaldurssonStefanMSc[1].pdf

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BLDC Motor Modelling and Control – A Matlab®/Simulink® 
Implementation 
 
 
 
 
 
 
 
 
 
– Master Thesis work by Stefán Baldursson – 
 
May, 2005 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Institutionen för Energi och Miljö 
International masters program in Electric Power Engineering 
CHALMERS TEKNISKA HÖGSKOLA 
Göteborg, Sverige, 2005 
Examinator: Torbjörn Thiringer 
 
 

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1
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v12 = −Vs
v23 =
0
v31 = Vs
i3 = 0⇒


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v12 = −Vs
v23 =
1
2
(Vs + e1 + e2 − 2e3)
v31 =
1
2
(Vs − e1 − e2 + 2e3)
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u12 = v12 − (e1 − e2) =
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u23 = v23 − (e2 − e3) =
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u31 = v31 − (e3 − e1) = −Vs − e3 + e1
i3 = 0⇒
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Vs − e1 + e2
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1
2
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u31 = v31 − (e3 − e1) =
1
2
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60◦ − 120◦
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180◦ − 240◦
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ia = 0
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2
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−Vs − eb + ec
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B
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D2
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A
B
C
D3
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60o - 120o
D5
D2
D1
D4
D5
B
B
B
A
A
A
C
C
C
C
A
B
120o - 180o
180o - 240o
240o - 300o
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6= 0
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ia
6= 0, ic = 0
−Vs − ea + eb
1
2
(Vs + ea − eb)
ia = 0, ic
6= 0
1
2
(−Vs + eb − ec)
Vs − eb + ec
ia = 0, ic = 0
0
0
60◦ − 120◦
ic
6= 0, ib
6= 0
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ic
6= 0, ib = 0
1
2
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1
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ic = 0, ib
6= 0
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ic = 0, ib = 0
0
0
120◦ − 180◦
ib
6= 0, ia
6= 0
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ib
6= 0, ia = 0
1
2
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ib = 0, ia
6= 0
1
2
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1
2
(−Vs − ea + ec)
ib = 0, ia = 0
0
0
180◦ − 240◦
ia
6= 0, ic
6= 0
Vs − ea + eb
−Vs − eb + ec
ia
6= 0, ic = 0
Vs − ea + eb
1
2
(−Vs + ea − eb)
ia = 0, ic
6= 0
1
2
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ia = 0, ic = 0
0
0
240◦ − 300◦
ic
6= 0, ib
6= 0
Vs − ea + eb
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ic
6= 0, ib = 0
1
2
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1
2
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ic = 0, ib
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0
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6= 0
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1
2
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6= 0
1
2
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1
2
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0
0
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D4
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3
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4
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 t(s)
Mech. speed (rpm)

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Mech. speed (rpm)

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0.4
0.5
0.6
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Te (mNm)

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1.5
2
2.5
3
3.5
4
4.5
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−0.15
−0.1
−0.05
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0.05
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ic (A)

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2.5
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Mech. speed (rpm)

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0.2
0.3
0.4
0.5
0.6
0.7
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0.2
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0.6
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0.002
0.003
0.004
0.005
0.006
0.007
0.008
0.009
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2.5
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Mech. speed (rpm)

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