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November 10, 2005 6.131 Lecture 24 1
Massachusetts Institute of Technology
Department of Electrical Engineering and Computer Science
6.131 Power Electronics Laboratory
Lecture 24
November 10, 2005
November 10, 2005 6.131 Lecture 24 2
November 10, 2005 6.131 Lecture 24 3
November 10, 2005 6.131 Lecture 24 4
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November 10, 2005 6.131 Lecture 24 6
Coils are wound on a toroidal core ( a magnetic circuit)
November 10, 2005 6.131 Lecture 24 7
Axial View: Magnets are interacting with stator current
November 10, 2005 6.131 Lecture 24 8
Magnetic materials exhibit hysteresis
For core materials you want a narrow curve
But for PM materials you want a wide curve
November 10, 2005 6.131 Lecture 24 9
Hard permanent magnet materials have B-H curves like this
Remanent Flux density can be as high as 1.4 T
Incremental permeability is like free space
November 10, 2005 6.131 Lecture 24 10
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By = Br + μ0Hm = μ0Hg
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gHg + hmHm = 0
Magnet Characteristic
Geometry
Combination
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By = Brhmg+ hm
November 10, 2005 6.131 Lecture 24 11
This graphic shows that calculation
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s = −μ0hmg
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Nomenclature: Two more views of the machine
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This is a ‘cut’ from the radial direction (section BB)
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Here is a cut through the machine (section AA)
Winding goes around the core: looking at 1 turn
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Voltage is induced by motion and magnetic field
Induction is:
Voltage induction rule:
Note magnets must agree!
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E '= E + v × B
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V = ulB =ωRlB =Cω
November 10, 2005 6.131 Lecture 24 16
Single Phase Equivalent Circuit of the PM machine
Ea is induced (‘speed’) voltage
Inductance and resistance are as expected
This is just one phase of three
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Ea =ωλ 0Voltage relates to flux:
November 10, 2005 6.131 Lecture 24 17
PM Brushless DC Motor is a synchronous PM machine with an inverter:
November 10, 2005 6.131 Lecture 24 18
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Ea = E cos ωt +θ( )
Eb = E cos ωt +θ −2π
3
⎛
⎝ ⎜
⎞
⎠ ⎟
Ec = E cos ωt +θ +2π
3
⎛
⎝ ⎜
⎞
⎠ ⎟
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Ia = I1 cosωt
Ib = I1 cos ωt −2π
3
⎛
⎝ ⎜
⎞
⎠ ⎟
Ic = I1 cos ωt +2π
3
⎛
⎝ ⎜
⎞
⎠ ⎟
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P =1
2EI1 cosθ =
1
2λ 0ωI1 cosθ
Induced voltages are:
Assume we drive with balanced currents:
Then converted power is:
Torque must be:
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T =p
ωP =
p
2λ 0I1 cosθ
November 10, 2005 6.131 Lecture 24 19
Now look at it from the torque point of view:
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T = IlBR =CI
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I1 =4
πsin
120o
2I0 =
4
π
3
2I0
Terminal Currents look like this:
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T =3
2pλ 0I1 = p
3 3
πλ 0I0
So torque is, in terms of DC side current:
November 10, 2005 6.131 Lecture 24 21
Va VbVc
Vab
0 12
Rectified back voltage is max of all six line-line voltages
<Eb>
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< Eb >=3
π3
−π
6
π
6
∫ ωλ 0 cosωtdωt
=3 3
πωλ 0
Average Rectified Back Voltage is:
Power is simply:
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Pem =3 3
πωλ 0I0 =KI0
November 10, 2005 6.131 Lecture 24 23
So from the DC terminals this thing looks like the DC machine:
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T =KI
Ea =Kω
K = p3 3
πλ 0
November 10, 2005 6.131 Lecture 24 24
Magnets must match (north-north, south-south) for the two rotor disks.
Looking at them they should look like this:
End A End B
Keyway
November 10, 2005 6.131 Lecture 24 25
Need to sense position: Use a disk that looks like this
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Position sensor looks at the disk: 1=‘white, 0=‘black’
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Some care is required in connecting to the position sensor
Vcc
GND
Channel 1
Channel2
(you need to figure out which of these is ‘count’ and which is ‘zero’)
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Control Logic:
Replace open loop with position measurement