june 9, 20056.11s 20051 massachusetts institute of technology department of electrical engineering...
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June 9, 2005 6.11s 2005 1
Massachusetts Institute of Technology
Department of Electrical Engineering and Computer Science
6.11s Design of Motors, Generators and Drive Systems
June 9, 2005
J.L. Kirtley Jr.
©2005, J.L. Kirtley Jr.
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Coils are wound on a toroidal core ( a magnetic circuit)
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Axial View: Magnets are interacting with stator current
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Magnetic materials exhibit hysteresis
For core materials you want a narrow curve
But for PM materials you want a wide curve
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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
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€
By = Br + μ0Hm = μ0Hg
€
gHg + hmHm = 0
Magnet Characteristic
Geometry
Combination
€
By = Brhmg+ hm
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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!
€
E '= E + v × B
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V = ulB =ωRlB =Cω
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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
€
Ea =ωλ 0Voltage relates to flux:
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PM Brushless DC Motor is a synchronous PM machine with an inverter:
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€
Ea = E cos ωt +θ( )
Eb = E cos ωt +θ −2π
3
⎛
⎝ ⎜
⎞
⎠ ⎟
Ec = E cos ωt +θ +2π
3
⎛
⎝ ⎜
⎞
⎠ ⎟
€
Ia = I1 cosωt
Ib = I1 cos ωt −2π
3
⎛
⎝ ⎜
⎞
⎠ ⎟
Ic = I1 cos ωt +2π
3
⎛
⎝ ⎜
⎞
⎠ ⎟
€
P =1
2EI1 cosθ =
1
2λ 0ωI1 cosθ
Induced voltages are:
Assume we drive with balanced currents:
Then converted power is:
Torque must be:
€
T =p
ωP =
p
2λ 0I1 cosθ
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Now look at it from the torque point of view:
€
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:
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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:
€
Pem =3 3
πωλ 0I0 =KI0
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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
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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
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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