6.11s June 2006 L3 1

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 2

Cross Section View: Surface Magnet Machine: Note windings

6.11s June 2006 L3 3

Alternate: Surface Mount (‘Iron Free’) Armature Winding

6.11s June 2006 L3 4

Magnets Inside the Rotor

6.11s June 2006 L3 5

Machine Design for Very High (negative) Saliency

6.11s June 2006 L3 6

Focus on Rating:

€

P + jQ =q

2VI =

q

2

Ea

NNI

V

Ea

Ea =ωλ

N=ωφ

φ =2RlB1

pkw

Rating is number of phases times voltage times current

Internal voltage is frequency times flux

And flux is the integral of Flux density

We will consider winding factor below

6.11s June 2006 L3 7

Internal Voltage Construction: Here is flux Density from Magnets

€

B(θ) = Bn sin npθ( )n

∑

Bn =4

πsin

npθm

2

⎛

⎝ ⎜

⎞

⎠ ⎟kgnBr

kgn =hm

hm+g

This is an approximation to the shape of the field in the air-gap (only an approximation)

Radial field

But see the notes for this done right

6.11s June 2006 L3 8

€

kg =Ri

p−1

Rs2p − Ri

2p

p

p+1R2

p+1 − R1p+1

( ) + Rs2p p

p−1R2

1− p − R11− p

( ) ⎛

⎝ ⎜

⎞

⎠ ⎟

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

€

kg =Rs

p−1

Rs2p − Ri

2p

p

p+1R2

p+1 − R1p+1

( ) + Ri2p p

p−1R2

1− p − R11− p

( ) ⎛

⎝ ⎜

⎞

⎠ ⎟

Stator winding outside:

Stator winding inside:

6.11s June 2006 L3 9

Current Capacity

€

NaIa = JslotAslot

Nslots

2q

NslotsAslot ≅ 2πRλ shs

Better :

NslotsAslot = π R + hs( )2

− R2( ) − Nshswt

Nswt = 2πR 1− λ s( )

This begs two questions:

How to establish current density?

How to establish slot fraction?

6.11s June 2006 L3 10

Voltage Ratio

€

Ea2 =V 2 + Xd Ia( )

2− 2VXd Ia cosθ

Ea2 =V 2 + Xd Ia( )

2+ 2VXd Ia sinψ

1 =V

Ea

⎛

⎝ ⎜

⎞

⎠ ⎟

2

+Xd IaEa

⎛

⎝ ⎜

⎞

⎠ ⎟

2

+ 2V

Ea

Xd IaEa

sinψ

V

Ea

= 1−Xd IaEa

cosψ ⎛

⎝ ⎜

⎞

⎠ ⎟

2

−Xd IaEa

sinψ

6.11s June 2006 L3 11

Calculation of Inductance: Start with a Full-Pitch Coil Set

This current distribution makes the flux distribution below

6.11s June 2006 L3 12

€

Br = μ0

4

π

NaIapgm

sin pθ( )

λ = lNa Br θ( )Rdθ0

π

p

∫

La = μ0

4

π

RlNa

p2gm

Ld = μ0

3

2

4

π

RlNakw2

p2gm

Ld = μ0

3

2

4

π

RlNakw2

p2(g+ hm )

Fundamental Flux Density

Flux Linkage

Idealized inductance of a full-pitch coil

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

6.11s June 2006 L3 13

€

kw = kbkp

This is what we mean by short pitch: see the original drawing

€

λ fp = l B1 sin θ( )Rdθ0

π

p

∫ =2RlB1

p

λ sp = l B1 sin θ( )Rdθπ

2p−

α

2p

π

2p+

α

2p

∫

=2RlB1

psin

α

2

kp =λ sp

λ fp

= sinα

2

6.11s June 2006 L3 14

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:

€

kb =sin

mγ

2

msinγ

2

The breadth factor is just the length of the addition of the vectors divided by the length of one times the number of vectors

6.11s June 2006 L3 15

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:

€

λ =℘sNc2Ia℘ s = μ0

hd

wd

+1

3

ws

hs

⎛

⎝ ⎜

⎞

⎠ ⎟

Llc = Nc2℘ s

Use top of slot dimensions for tapered slots: very small error

6.11s June 2006 L3 16

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 × 2p(m − Nsp ) + Llc × 2pNsp

= Llc 8pm − 6pNsp( )

6.11s June 2006 L3 17

Winding resistance is important

€

Ra = ρl w

Aw

So there are various ways of estimating winding length and area: Area is easier:

€

Aw = λ a

As

2Nc

Nc =Na

2mp

€

l w = Na 2l + 2l e( )Winding length must account for end turns and that is a geometric problem

6.11s June 2006 L3 18

We have power conversion figure out

Losses are:

Armature conduction loss: I2 Ra

Core Loss

Friction, windage, etc

€

Bc = B1

R

pd c

Bt =B1

1− λ s( )

To get core loss we use the model developed earlier, depending on the species of iron and fields calculated thus:

6.11s June 2006 L3 19

The Process of design is a loop

6.11s June 2006 L3 20

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

6.11s June 2006 L3 21

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

6.11s June 2006 L3 22

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)

6.11s June 2006 L3 23

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

6.11s June 2006 L3 24

Then per-unit torque is:

Per-Unit Currents to achieve the maximum torque per unit current are:

6.11s June 2006 L3 25

Note that per-unit flux achievable for a given terminal voltage is:

And this is related to current by:

6.11s June 2006 L3 26

6.11s June 2006 L3 27

6.11s June 2006 L3 28

Base speed

6.11s June 2006 L3 29

Here is your basic three phase bridge

6.11s June 2006 L3 30

Suppose we have this situation:

6.11s June 2006 L3 31

Here is one way of switching that circuit:

The arrows designate when a switch is ON

6.11s June 2006 L3 32

Here is what is on in State 0:

Va = V, Vb = V, Vc = 0 Vn = 2V/3

6.11s June 2006 L3 33

Va = 0, Vb = V, Vc = 0 Vn = V/3

Here is what is on in State 1:

6.11s June 2006 L3 34

Va = 0, Vb = V, Vc = V Vn = 2V/3

Here is what is on in State 2:

6.11s June 2006 L3 35

Va = 0, Vb = 0, Vc = V Vn = V/3

Here is what is on in State 3:

6.11s June 2006 L3 36

Va = V, Vb = 0, Vc = V Vn = 2V/3

Here is what is on in State 4:

6.11s June 2006 L3 37

Va = V, Vb = 0, Vc = 0 Vn = V/3

Here is what is on in State 5:

6.11s June 2006 L3 38

Voltages: Line-Line Voltages are well defined

6.11s June 2006 L3 39

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

6.11s June 2006 L3 40

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)

6.11s June 2006 L3 41

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

6.11s June 2006 L3 42

Here is a comparison of switching strategies

180 degree six-pulse

120 degree six pulse

Give up a little timing between switch closings

6.11s June 2006 L3 43

Switches Q_1 and Q_5 are on: State0

Va = V, Vb = 0, Vc = V/2

6.11s June 2006 L3 44

Switches Q_1 and Q_6 are on: State1

Va = V, Vc = 0, Vb = V/2

6.11s June 2006 L3 45

Switches Q_2 and Q_6 are on: State2

Vb = V, Vc = 0, Va = V/2

6.11s June 2006 L3 46

Switches Q_2 and Q_4 are on: State3

Va = 0, Vb = V, Vc = V/2

6.11s June 2006 L3 47

Switches Q_3 and Q_4 are on: State4

Va = 0, Vc = V, Vb = V/2

6.11s June 2006 L3 48

Switches Q_3 and Q_5 are on: State5

Vc = V, Vb = 0, Va = V/2

6.11s June 2006 L3 49

This switching pattern results in these voltages

6.11s June 2006 L3 50

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

6.11s June 2006 L3 51

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

6.11s June 2006 L3 52

NAND (Not AND)

Is the same as Negative Input OR

The ‘138 output is ‘active low’:

Matching bubbles makes an OR function

6.11s June 2006 L3 53

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’

6.11s June 2006 L3 54

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

6.11s June 2006 L3 55

And here are the counter states: note how CL works

6.11s June 2006 L3 56

We already detect state 5 with the ‘138

6.11s June 2006 L3 57

The ‘138 is a simple selector: use like this:

And here are the pinouts of the ‘163 and ‘138

6.11s June 2006 L3 58

Variable Voltage: do the Pulse Width Modulation thing

6.11s June 2006 L3 59

Nomenclature: Two more views of the machine

6.11s June 2006 L3 60

This is a ‘cut’ from the radial direction (section BB)

6.11s June 2006 L3 61

Here is a cut through the machine (section AA)

Winding goes around the core: looking at 1 turn

6.11s June 2006 L3 62

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

6.11s June 2006 L3 63

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:

6.11s June 2006 L3 64

PM Brushless DC Motor is a synchronous PM machine with an inverter:

6.11s June 2006 L3 65

€

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

6.11s June 2006 L3 66

Now look at it from the torque point of view:

€

T = IlBR = CI

6.11s June 2006 L3 67

€

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:

6.11s June 2006 L3 68

Va VbVc

Vab

0 12

Rectified back voltage is max of all six line-line voltages

<Eb>

6.11s June 2006 L3 69

€

< 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

6.11s June 2006 L3 70

So from the DC terminals this thing looks like the DC machine:

€

T = KI

Ea = Kω

K = p3 3

πλ 0

6.11s June 2006 L3 71

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

6.11s June 2006 L3 72

Need to sense position: Use a disk that looks like this

6.11s June 2006 L3 73

Position sensor looks at the disk: 1=‘white, 0=‘black’

6.11s June 2006 L3 74

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’)

6.11s June 2006 L3 75

Control Logic:

Replace open loop with position measurement

6.11s June 2006 L3 76

Why do we need to PWM only the top switches?

What happens with you turn OFF switch Q1?