Transcript 6.11s Notes for Lecture 3 June 14, 2006 J.L. Kirtley Jr.

6.11s Notes for Lecture 3
PM ‘Brushless DC’ Machines: Elements of Design
June 14, 2006
J.L. Kirtley Jr.
6.11s June 2006 L3
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Cross Section View: Surface Magnet Machine: Note windings
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Alternate: Surface Mount (‘Iron Free’) Armature Winding
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Magnets Inside the Rotor
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Machine Design for Very High (negative) Saliency
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Focus on Rating:
q
q Ea
V
P  jQ  VI 
NI
2
2 N
Ea
Ea  

 
Rating is number of
phases times voltage
times current
Internal voltage is
frequency times flux
N
2R B1

kw
p
And flux is the integral of
Flux density
We will consider winding
factor below

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Internal Voltage Construction: Here is flux Density from Magnets
This is an approximation to
the shape of the field in the
air-gap (only an
approximation)
B( )   Bn sin np 
Radial field
n
np m 
Bn  sin 
k gn Br
  2 
hm
kgn 
hm g
4
But see the notes for this
done right
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Magnetic field can be found through a little field analysis
The result below is
good for magnets
inside and p not
equal to one. See
the notes for other
expressions
Stator winding outside:

Rsp1  p
p
p 1
p 1
2p
1 p
1 p
kg  2 p
R2  R1  Ri
R2  R1 


2 p 
Rs  Ri p  1
p 1

Stator winding inside:


Rip1  p
p
p 1
p 1
2p
1 p
1 p
kg  2 p
R2  R1  Rs
R2  R1 


2 p 
Rs  Ri p  1
p 1

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Current Capacity
N slots
N a Ia  J slot Aslot
2q
N slotsAslot  2Rshs
Better :


N slotsAslot   R  hs   R 2  N shsw t
2
N sw t  2R1 s 
This begs two questions:

How to establish current density?
How to establish slot fraction?
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Voltage Ratio
E  V  X d Ia   2VX d Ia cos
2
a
2
2
E  V  X d Ia   2VX d Ia sin 
2
a
2
2
 V 2 X d Ia 2
V X d Ia
1     
sin 
  2
Ea Ea
E a   E a 
2


V
X I
X I
 1  d a cos   d a sin 
Ea
Ea
 E a

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Calculation of Inductance: Start with a Full-Pitch Coil Set
This current
distribution makes
the flux distribution
below
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4 N a Ia
Br  0
sin p 
 pgm
Fundamental Flux Density

p
  N a  Br  Rd 
0
La   0
Idealized inductance of a fullpitch coil
4 R Na
 p 2 gm
3 4 R N a kw2
Ld   0
2  p 2 gm
3 4 R N a k w2
Ld   0
2  p 2 (g  hm )

Flux Linkage
Taking into account phase-phase
coupling (for 3 phase machine)
and winding factor
And for the PM machine the
magneti is part of the magnetic
gap
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kw  k b k p

 fp 
This is what we mean by
short pitch: see the
original drawing
p
 B1 sin  Rd 
0

sp 

2R B1
p

2p 2p

 B sin  Rd


1
2p 2p

2R B1


sin
p
2
sp

kp 
 sin
 fp
2

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Breadth Factor: Coils link flux slightly out of phase
Here is a construction of the
flux addition. It takes a bit of
high-school like geometry to
show that:
The breadth factor is just the
length of the addition of the
vectors divided by the length of
one times the number of vectors
m
sin
2
kb 

msin
2

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Slot Leakage:
Suppose the slot were to look
like this: It actually has two
coils that have Nc half turns
each.
Flux linked by one coil from
one driven coil is:
  sN c2 Ia
hd 1 w s 
s  0 

w
3
h
 d
s 
Llc  N c2s

Use top of slot dimensions for
tapered slots: very small error
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There are 2p(m-Nsp) slots with both coils in the same phase
And 2p Nsp slots with coils ineach of the different phases (in
each phase)
So slot leakage is
Lsl  4Llc  2 p(m  N sp )  Llc  2 pN sp
 Llc 8 pm  6 pN sp 

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Winding resistance is important
Ra  
Aw
As
Aw  a
2N c
So there are various ways of
estimating winding length and area:
Area is easier:

Winding length must account
for end turns and that is a
geometric problem
w
Na
Nc 
2mp
w
 N a 2  2
e



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We have power conversion figure out
Losses are:
Armature conduction loss: I2 Ra
Core Loss
Friction, windage, etc
To get core loss we use the
model developed earlier,
depending on the species of
iron and fields calculated
thus:
R
Bc  B1
pd c
Bt 
B1
1 s 

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The Process of design is a loop
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There are (at least) three types of performance specifications:
Requirements are specifications that must be met
a. Rotational Speed or frequency
b. Rating
Limits are specifications that must not be exceeded
a. Tip Speed
b. Maximum operating temperature
Attributes are specifications that, all other things being
equal, should be maximized or minimized
So the design process consists of meeting the
requirements, observing the limits and maximizing the
attributes
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Multiple attributes make maximization iffy
No simple way of telling if A is
better than D (or C)
But B is clearly
superior to
(dominates) E
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Novice Design Assistant:
Is deliberately not an expert system
Uses Monte Carlo to generate randomized designs
Each variable in the design space is characterized by:
Mean Value
Standard Deviation
Maximum value (limit)
Minimum value (limit)
Setup file (msetup.m) specifies
Number of design variables
For each: the above data
Number of attributes to be returned
function file called by nda.m: called attribut.m
returns attributes and a go-no-go (limits not violated)
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Operation
For the PM machine fluxes are
given by simple expressions
So torque is:
Now normalize the machine
in the following way: probably
use field flux for
normalization
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Then per-unit torque
is:
Per-Unit Currents to achieve the maximum torque per unit current are:
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Note that per-unit flux achievable for a
given terminal voltage is:
And this is related to current by:
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Base speed
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Here is your basic three phase bridge
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Suppose we have this situation:
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Here is one way of switching that circuit:
The arrows designate when a switch is ON
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Here is what is on in State 0:
Va = V, Vb = V, Vc = 0 Vn = 2V/3
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Here is what is on in State 1:
Va = 0, Vb = V, Vc = 0 Vn = V/3
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Here is what is on in State 2:
Va = 0, Vb = V, Vc = V Vn = 2V/3
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Here is what is on in State 3:
Va = 0, Vb = 0, Vc = V Vn = V/3
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Here is what is on in State 4:
Va = V, Vb = 0, Vc = V Vn = 2V/3
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Here is what is on in State 5:
Va = V, Vb = 0, Vc = 0 Vn = V/3
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Voltages: Line-Line Voltages are well defined
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To generate switching signals:
•Totem Pole A is High in states 0, 4 and 5
•Totem Pole B is High in states 0, 1 and 2
•Totem Pole C is High in states 2, 3 and 4
This allows us to use very simple logic:
A = S0 + S4 + S5
B = S0 + S1 + S2
C = S2 + S3 + S4
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To generate switch signals
Note that either top or bottom switch is on in each phase
Generation of states: we will do this a bit later (see below)
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This ‘six pulse’ switching strategy:
•Makes good use of the switching devices
•Also requires ‘shoot-through’ delays
•Has very simple logic
We propose an alternative switching strategy
•Makes minimally less effective use of switches
•Uses a little more logic
•But does not risk shoot through
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Here is a comparison of switching strategies
180 degree sixpulse
120 degree six
pulse
Give up a little
timing between
switch closings
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Switches Q_1 and Q_5 are on: State0
Va = V, Vb = 0, Vc = V/2
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Switches Q_1 and Q_6 are on: State1
Va = V, Vc = 0, Vb = V/2
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Switches Q_2 and Q_6 are on: State2
Vb = V, Vc = 0, Va = V/2
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Switches Q_2 and Q_4 are on: State3
Va = 0, Vb = V, Vc = V/2
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Switches Q_3 and Q_4 are on: State4
Va = 0, Vc = V, Vb = V/2
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Switches Q_3 and Q_5 are on: State5
Vc = V, Vb = 0, Va = V/2
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This switching pattern results in these voltages
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Switches turn on:
Q1
State_0 OR State_1
Q2
State_2 OR State_3
Q3
State_4 OR State_5
Q4
State_3 OR State_4
Q5
State_1 OR State_5
Q6
State_1 OR State_2
Each switch is on for two states
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So here is how to do it
3 bit input to ‘138 selects one of 8 outputs
Active low output!
‘138 has 3 enable inputs: two low, one high
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NAND (Not AND)
Is the same as Negative Input OR
The ‘138 output is
‘active low’:
Matching bubbles
makes an OR function
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Now we must generate six states in sequence
If we have a ‘clock’ with rising edges at the right
time interval we can use a very simple finite state
machine
This could be a counter, reset when it sees ‘5’
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Here is a good counter to use: 74LS163
This is a loadable counter: don’t need that feature
Clear function is synchronous: so it clears only ON a
clock edge
Part is ‘edge triggered’: changes state on a positive
clock edge
P and T are enables: must pull them high
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And here are the counter states: note how CL works
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We already detect state 5 with the ‘138
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The ‘138 is a simple selector: use like this:
And here are the pinouts of the ‘163 and ‘138
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Variable Voltage: do the Pulse Width Modulation thing
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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:
E' E  v  B
Voltage induction rule:
V  u B  R B  C
Note magnets must agree!


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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
Voltage relates to flux:
Ea  0
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PM Brushless DC Motor is a synchronous PM
machine with an inverter:
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Induced voltages
are:
Assume we drive with
balanced currents:
Ia  I1 cos t

2 
Ib  I1 cost  

3 

2 
Ic  I1 cost 


3 
E a  E cost   

2 
E b  E cost    

3 

2 
E c  E cost   


3 
Then converted power is:


Torque must be:

1
1
P  EI1 cos  0I1 cos
2
2
p
p
T  P  0I1 cos


2
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Now look at it from the torque point of view:
T  I BR  CI

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Terminal Currents look like this:
I1 
4

sin
120
4 3
I0 
I0
2
 2
So torque is, in terms of DC side current:

3
3 3
T  p0 I1  p
0 I 0
2


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Rectified back voltage is max of all six line-line voltages
Va
Vc
Vb
0
12
Vab
<Eb>
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Average Rectified Back Voltage is:

 E b 

3 3

6

3



30 cos tdt
6
0
Power is simply:
Pem 
3 3

0 I0  KI0


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So from the DC terminals this thing looks like the DC
machine:
T  KI
E a  K
Kp
3 3

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
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Why do we need to PWM only the top switches?
What happens with you turn OFF switch Q1?
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