June 9, 20056.11s 20051 Massachusetts Institute of Technology Department of Electrical Engineering...
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![Page 1: June 9, 20056.11s 20051 Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.11s Design of Motors, Generators.](https://reader035.fdocuments.us/reader035/viewer/2022062421/56649e565503460f94b4e0fa/html5/thumbnails/1.jpg)
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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June 9, 2005 6.11s 2005 2
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June 9, 2005 6.11s 2005 3
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Coils are wound on a toroidal core ( a magnetic circuit)
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June 9, 2005 6.11s 2005 7
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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June 9, 2005 6.11s 2005 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
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June 9, 2005 6.11s 2005 10
€
By = Br + μ0Hm = μ0Hg
€
gHg + hmHm = 0
Magnet Characteristic
Geometry
Combination
€
By = Brhmg+ hm
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June 9, 2005 6.11s 2005 11
This graphic shows that calculation
€
s = −μ0hmg
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June 9, 2005 6.11s 2005 12
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
€
V = ulB =ωRlB =Cω
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June 9, 2005 6.11s 2005 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
€
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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June 9, 2005 6.11s 2005 19
Now look at it from the torque point of view:
€
T = IlBR =CI
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June 9, 2005 6.11s 2005 20
€
I1 =4
πsin
120o
2I0 =
4
π
3
2I0
Terminal Currents look like this:
€
T =3
2pλ 0I1 = p
3 3
πλ 0I0
So torque is, in terms of DC side current:
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June 9, 2005 6.11s 2005 21
Va VbVc
Vab
0 12
Rectified back voltage is max of all six line-line voltages
<Eb>
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June 9, 2005 6.11s 2005 22
€
< 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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June 9, 2005 6.11s 2005 23
So from the DC terminals this thing looks like the DC machine:
€
T =KI
Ea =Kω
K = p3 3
πλ 0
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June 9, 2005 6.11s 2005 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
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Need to sense position: Use a disk that looks like this
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June 9, 2005 6.11s 2005 26
Position sensor looks at the disk: 1=‘white, 0=‘black’
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June 9, 2005 6.11s 2005 27
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