S.D. Sudhoff Fall 2005 - Purdue University
Transcript of S.D. Sudhoff Fall 2005 - Purdue University
EE595S: Class Lecture NotesChapter 15: Brushless DC Motor Drives
S.D. Sudhoff
Fall 2005
Fall 2005 EE595S Electric Drive Systems 2
15.1 Architecture
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15.2 Topology
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15.3 Equivalence of VSI Schemesto Idealized Source
• 180 VSI• Duty Cycle Modulation• Sine-Triangle Modulation• Sine-Triangle Modulation (with 3rd)• Space-Vector Modulation
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180 VSI Operation
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180 VSI Operation
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180 VSI Operation
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180 VSI Operation
• Comparing the two figures(15.3-1)
• Now recall(15.3-2)
(15.3-3)
hrc φθθ +=
dccqs vv π
2=
0=cdsv
Fall 2005 EE595S Electric Drive Systems 9
180 VSI Operation
• Using the frame-to-frame transformation(15.3-4)(15.3-5)
• Thus(15.3-6)(15.3-7)
hdcrqs vv φπ cos2=
hdcrds vv φπ sin2−=
dcs vvπ2
21
=
hv φφ =
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180 VSI Operation
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Duty Cycle Modulation
• In this case(15.3-8)(15.3-9)
• Thus(15.3-10)(15.3-11)
• So(15.3-12)
(15.3-7)
dccqs vdv π
2=
0=cdsv
hdcrqs vdv φπ cos2=
hdcrds vdv φπ sin2−=
dcs vdvπ2
21
=
hv φφ =
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Duty Cycle Modulation
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Sine-Triangle Modulation
• In this case, converter angle calculated as(15.3-13)
• Now recall
(15.3-14)
(15.3-15)
(15.3-16)
vrc φθθ +=
⎪⎩
⎪⎨⎧
>
≤<=
1)(
102
21
ddfv
dvdv
dc
dccqs
π
0=cdsv
( ) 11arccos21)( 4121
21 >⎟
⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛−+−= d
dddf d π
Fall 2005 EE595S Electric Drive Systems 14
Sine-Triangle Modulation
• Converting to the rotor reference frame
(15.3-17)
• (15.3-18)
⎪⎩
⎪⎨⎧
>
≤=
1cos)(
1cos2
21
ddfv
ddvv
vdc
vdcrqs φ
φ
π
⎪⎩
⎪⎨⎧
>−
≤−=
1sin)(
1sin2
21
ddfv
ddvv
vdc
vdcrds φ
φ
π
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Sine-Triangle Modulation
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Sine Triangle Modulationwith 3rd Harmonic Injection
• Following our earlier work, we will have(15.3-19)(15.3-20)
3/20cos21 ≤≤= dvdv vdc
rqs φ
3/20sin21 ≤≤−= dvdv vdc
rds φ
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Space Vector Modulation
• In this case we have
(15.3-21)
(15.3-22)
(15.3-23)
⎪⎪⎩
⎪⎪⎨
⎧
≥
<
=3/
3
3/
**
*
**
dcspkspk
rqsdc
dcspkrqs
rqs vv
v
vv
vvv
v
⎪⎪⎩
⎪⎪⎨
⎧
≥
<
=3/
3
3/
**
*
**
dcspkspk
rdsdc
dcspkrds
rds vv
v
vv
vvv
v
( ) ( )2*2** rds
rqsspk vvv +=
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Modulation Summary
• For all VSI strategies, we have(15.3-24)(15.3-25)
• Where
(15.3-26)
vdcrqs mvv φcos=
vdcrds mvv φsin−=
( )( )⎪
⎪⎪⎪⎪⎪
⎩
⎪⎪⎪⎪⎪⎪
⎨
⎧
≥−
≤−
≤
≥
≤
=
ο
3/vvs
3/vvvv
dd
ddf
dd
d
m
dc*spk
dc*spk
dc
spk
modulationvectorpace3
1
modulationvectorspace
)3/2( modulation harmonic-third
1) (modulation triangle-sine)(
1)( modulation triangle-sine
modulation cycle-duty
modulation source- voltage180
*21
221
2
2
π
π
π
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Modulation Summary
• Note, for space vector modulation(15.3-27))( ** r
dsrqsv jvvangle −=φ
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15.4 Average-Value Analysis of VSI Drives
• Average-value rectifier model(15.4-1)
• Where(15.4-2)
(15.4-3)
(15.4-4)
rrrrror iplirvv −−= αcos
rectifierphasesingle rectifierphase three
2
22
330
⎪⎩
⎪⎨⎧
=E
Evr
π
π
rectifierphase single rectifierphase three
2
3
⎪⎩
⎪⎨⎧
=ceu
ceur L
Lr
ω
ω
π
π
rectifierphase single rectifierphase three2
⎩⎨⎧
=c
cr L
Ll
Fall 2005 EE595S Electric Drive Systems 21
Working Out the Details…
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Final Average Value Model
• Nonlinear Average Value Model (NLAM)
(15.4-21)( )
⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−−+−
−
−−
+
⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
′−
−−
−−
−−−
=
⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
))((sin
cos00
)sin(coscos
000000000000
00000
00000000000000000000
2231
2
1
1
23
01
1222
3
'
1
11
11
1
Lr
dsr
qsqdP
JP
rqsrL
LvdcL
rdsrL
LvdcL
rdsv
rqsvC
mrL
r
rds
rqs
st
st
dc
r
mJP
Lr
q
mLr
C
LLr
L
CC
LLr
r
rds
rqs
st
st
dc
r
TiiLLimv
imv
iiv
iivivi
Liivivi
p
d
q
d
q
d
q
dc
rl
d
s
q
s
st
stst
st
st
dcdc
rlrl
rl
ωφ
ωφ
φφα
ωλ
λ
ω
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NLAM Startup PredictionFigure 15.6-2
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Detailed Startup PredictionFigure 15.6-1
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Steady-State Performance of VSI Drives
• For Steady State Conditions(15.5-1)
• From Our Earlier Work(15.4-14)
• Thus(15.5-2)
• The Steady-State Stabilizing Current is Zero So
(15.5-3)
rrldcr IrVv −−= αcos0 0
)sincos(23
vrdsv
rqsdc iimi φφ −=
)sincos(0 23
vrdsv
rqsstr IImII φφ −−−=
)sincos(0 23
vrdsv
rqsr IImI φφ −−=
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Steady-State Performance of VSI Drives
• Combining (15.5-3) with (15.5-1) yields(15.5-4)
• From the Machine Voltage Equations(15.5-5)(15.5-6)
)sincos(cos 23
0 vrdsv
rqsrlrdc IImrvV φφα −−=
mrrdsdr
rqssvdc ILIrmV λωωφ ′−−−= cos0
rqsqr
rdssvdc ILIrmV ωφ +−−= sin0
Fall 2005 EE595S Electric Drive Systems 27
Steady-State Performance of VSI Drives
• Solving (15.5-5) and (15.5-6) for the Currents
(15.5-7)
(15.5-8)
qdrs
vdcdrmrvdcsrqs
LLrmVLmVrI 22
sin)cos(ω
φωλωφ+
+′−=
qdrs
vdcsmrvdcqrrds LLr
mVrmVLI 22
sin)cos(
ω
φλωφω
+
−′−=
Fall 2005 EE595S Electric Drive Systems 28
Steady-State Performance of VSI Drives
• Substitution of (15.5-7) and (15.5-8) into (15.5-4) and Solving Yields
(15.5-9)
• Caution: This Expression Only Valid if DC Current (15.5-14) Is Positive
vqdrrlsrlqdrs
vqrvsmrrlrqdrsdc
mLLrrrmLLr
LrmrvLLrV
φωω
φωφλωαω
2sin)(
)sincos(cos)(2
432
2322
23
022
−+++
−′++=
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Steady-State Performance of VSI Drives
• Suppose the DC Current from (15.5-14) is Negative
This Can’t HappenThus
• (15.5-10)
Substitution of (15.5-7) and (15.5-8) Into (15.5-10) Yields
• (15.5-11)
0sincos =− vrdsv
rqs II φφ
)2sin)((
)sincos(
210
vqdrs
vqrvsmrIdc LLrm
LrV dc φω
φωφλω
−+
−′==
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Steady-State Performance of VSI Drives
• Solution Algorithm1. Calculate DC Voltage Using (15.5-9)2. Calculate QD Currents Using (15.5-7); (15.5-
8)3. Calculate DC Current Using (15.5-14)4. If DC Current >= 0, Goto 75. Calculate DC Voltage Using (15.5-11)6. Goto 27. Done
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Steady-State Performance of VSI Drives
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15.6 Transient Performance of VSI Drives
• Linearizing (15.4-21) We Have
(15.6-1)
[ ]
( ) ( )
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
∆∆∆∆∆
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−−
−
−
+
⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
∆∆
∆∆∆∆∆
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−−+′
−+−
−−−−
−−
−−
−−
=∆∆∆∆∆∆∆
+−
′
L
v
ro
JP
vLmv
vLv
vLmv
vLv
Ciim
Cii
LLv
r
rds
rqs
st
st
dc
r
rqsqdJ
PrdsqdmJ
P
rqsL
LLr
rLL
vLm
LrdsL
LrL
LLr
vLm
Cst
LstLstr
Lst
vCm
vCm
CC
LLr
Tr
rds
rqsststdcr
T
mv
iivivi
iLLiLL
i
i
iivivip
d
dc
d
dc
q
dc
q
dc
dc
rdsv
rqsv
dc
rdsv
rqsv
rlrl
r
d
q
d
s
d
q
d
q
m
q
d
q
d
q
s
q
stdcdcdcdc
rlrl
rl
φ
α
φφ
φφ
ωλ
ωφ
ωφ
φφ
ω
φφφφ
α
λ
cos
0000
0cossin00
0sincos000000000000000
000
0)())((0000
00sin0
00cos0
0000000000
0sincos00
00000
12
00
00
)cos(sin23)sin(cos
23
cos
012
223
012
223
000
000
1
11
023
02311
1
000
000
000000000
00
0
0
00
Fall 2005 EE595S Electric Drive Systems 33
Transient Performance of VSI Drives
• In (15.6-1), Note that(15.6-2)xxx ∆+= 0
Fall 2005 EE595S Electric Drive Systems 34
Predictions of Linearized ModelStartup
Fall 2005 EE595S Electric Drive Systems 35
Predictions of Detailed ModelStartup
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Predictions of Linearized ModelStep Change in Duty Cycle
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Predictions of Detailed ModelStep Change in Duty Cycle
Fall 2005 EE595S Electric Drive Systems 38
15.8 Case Study: VSI Based Speed Control
• Parameters
• Design ObjectivesLoad is InertialNo Steady-State ErrorPhase Margin of 60o at Nominal Speed of 200 rad/s(mechanical)
E 85.5 V
euω 377 rad/s
cL 5 mH
dcL 5 mH
dcC 1000 Fµ
J 5 mN!m!s2
sr 2.98Ω
qL 11.4 mH
dL 11.4 mH
mλ ′ 0.156 Vs
P 4
Fall 2005 EE595S Electric Drive Systems 39
Control Law
• Mathematically
(15.8-1)∫ −+−= ∗∗ dtKd rmrmK
rmrm )()( ωωωω τ
Fall 2005 EE595S Electric Drive Systems 40
Starting Point: Open Loop Frequency Response
• ObserveGain Margin is InfinitePhase Margin is 20o
Fall 2005 EE595S Electric Drive Systems 41
Selection of τ
• Integral Feedback Will Decrease Phase By 90o at Frequencies Less Than 1/(2πτ)
• We Will Select Breakpoint Frequency at 0.01 Hz for τ of 16 s
Fall 2005 EE595S Electric Drive Systems 42
Compensated Frequency Response
Fall 2005 EE595S Electric Drive Systems 43
Selection of K
• Select K to Achieve Phase Margin of 60o
• Thus, We Choose K=0.25 (-12 dB)
Fall 2005 EE595S Electric Drive Systems 44
Compensated Frequency Response
Fall 2005 EE595S Electric Drive Systems 45
Time-Domain Response
Fall 2005 EE595S Electric Drive Systems 46
Critique of Design
• Drive Becomes Quite Overmodulated• Bandwidth is Not Nearly 100 Hz• Significant Overcurrent (Rated is 2.6 A,
rms)• The Speed is not 200 rad/s At End of Study
Fall 2005 EE595S Electric Drive Systems 47
15.9 Current-Regulated Drives
• Current Based Inverter Current RegulationAdvantages
• Torque Control Bandwidth• Robustness w.r.t. Machine Parameters• Robust w.r.t. Faults• Control design with lower order system
Disadvantage• Switching Frequency Cannot Be Directly Controlled
Fall 2005 EE595S Electric Drive Systems 48
Hysteresis Current-Regulated Drive
Fall 2005 EE595S Electric Drive Systems 49
Performance of HysteresisControlled Drive
Fall 2005 EE595S Electric Drive Systems 50
Voltage Source Based Current Regulation
• AdvantagesControlled Switching FrequencyCan Offer Excellent Waveform Quality
• DisadvantageLower Control Bandwidth
Fall 2005 EE595S Electric Drive Systems 51
Sine Triangle Current Regulator
Fall 2005 EE595S Electric Drive Systems 52
Example 15B
• Suppose We Utilize
(15.B-1)
(15.B-2)
• Place Poles At s=-200, s=-1000
)()( rqs
rqs
ipm
rdsssr
rqs ii
sKKiLv −⎟
⎠⎞
⎜⎝⎛ ++′+= ∗∗ λω
)( rds
rds
ip
rqsssr
rds ii
sKKiLv −⎟
⎠⎞
⎜⎝⎛ ++−= ∗∗ ω
Fall 2005 EE595S Electric Drive Systems 53
Example 15B
• Results(15B-4)(15B-5)
sKi /2280Ω=
Ω= 7.10pK
Fall 2005 EE595S Electric Drive Systems 54
Performance During Step Change In Current
Fall 2005 EE595S Electric Drive Systems 55
15.10 Voltage Limitations of Current-Source Inverter Drives
• Lets Assume(15.10-1)(15.10-2)
• Then(15.10-3)
(15.10-4)
∗= rqs
rqs ii
∗= rds
rds ii
mrrdsdr
rqss
rqs iLirv λωω ′++= **
** rqsqr
rdss
rds iLirv ω−=
Fall 2005 EE595S Electric Drive Systems 56
15.10 Voltage Limitations of Current-Source Inverter Drives
• Recall That(15.10-5)
• Thus(15.10-6)
• Also Recall(15.10-7) (w/o Harmonics)
(15.10-8) (w Harmonics)
( ) ( )22
21 r
dsrqss vvv +=
22 )()(2
1 ∗∗∗∗ −+′++= rqsqr
rdssrm
rdsdr
rqsss iLiriLirv ωωλω
dcs vv6
1=
dcs vvπ2
=
Fall 2005 EE595S Electric Drive Systems 57
Response to Step Decrease in DC Voltage
Fall 2005 EE595S Electric Drive Systems 58
15.11 Current Command Synthesis(Non-Salient Machines)
• Recall (15.11-1)
• Thus(15.11-2)(15.11-3)
• Advantages
• But..
rqsm
Pe iT λ′= 22
3
∗′
∗ = ePrqs Ti
mλ12
32
0=∗rdsi
Fall 2005 EE595S Electric Drive Systems 59
15.11 Current Command Synthesis(Non-Salient Machines)
• Voltage Requirement(15.11-4)
• If Not Enough Voltage
(15.11-5)
(15.11-6)• This is Referred to as Flux Weakening• Warning:
22 )()(2
1 ∗∗∗∗ −+′++= rqsssr
rdssrm
rdsssr
rqsss iLiriLirv ωωλω
2
22222 )(2
z
izrvzLi
rqsmrssrssmr
ds
∗∗ +′−+′−=
λωωλ
222ssrs Lrz ω+=
Fall 2005 EE595S Electric Drive Systems 60
15.11 Current Command Synthesis(Salient Machines)
• Recall(15.11-7)
• Thus(15.11-8)
• Also Recall(15.11-9)
))((223 r
dsrqsqd
rqsm
Pe iiLLiT −+′= λ
qdm
rqsqdP
erds LLiLL
Ti−′
−−
= ∗∗ λ1
)(223
22 )()(2
1 ∗∗ += rds
rqss iii
Fall 2005 EE595S Electric Drive Systems 61
15.11 Current Command Synthesis(Salient Machines)
• Strategy: Minimize Fundamental Component of Current
• Why?
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15.11 Current Command Synthesis(Salient Machines)
• Result
• In this Case, What If Inadequate Voltage
Fall 2005 EE595S Electric Drive Systems 63
15.11 Current Command Synthesis(Salient Machines)
Fall 2005 EE595S Electric Drive Systems 64
15.12 Average-Value Modeling of Current Regulated Drives
• We Begin With(15.12-1)(15.12-2)
• Ignoring Stator Transients(15.12-3)(15.12-4)
• Thus(15.12-5)
*rqs
rqs ii =
∗= rds
rds ii
rmrdsdr
rqss
rqs iLirv ωλω ′++= ∗∗∗
∗∗ −= rqsqr
rdss
rds iLirv ω
])()([ 223 r
qsmrrds
rqsqdr
rds
rqss iiiLLiirP λωω ′+−++= ∗∗∗∗
Fall 2005 EE595S Electric Drive Systems 65
15.12 Average-Value Modeling of Current Regulated Drives
• Since(15.12-6)
• We Have(15.12-7)
• Again Assuming Our Currents Are Equal to Commanded Values
(15.12-8)
dcdc V
Pi =
])()([ 2123 r
qsmrrds
rqsqdr
rds
rqssVdc iiiLLiiri
dcλωω ′+−++= ∗∗∗∗
))((223 ∗∗∗ −+′= r
dsrqsqd
rqsm
Pe iiLLiT λ
Fall 2005 EE595S Electric Drive Systems 66
15.12 Average-Value Modeling of Current Regulated Drives
• Resulting State Model(15.12-9)
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−−+′
′+−++−+
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−−
−
−−
=
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
∗∗∗
∗∗∗∗∗
]))(([00
])()([
00000
0000
00
000
000
2231
2
21231
cos
1
11
11
1
0
Lrds
rqsqd
rqsm
PJ
P
rqsmr
rds
rqsqdr
rds
rqssVC
Lv
r
st
st
dc
r
C
LLr
L
CC
LLr
r
st
st
dc
r
TiiLLi
iiiLLiir
vivi
vivi
p
dcdc
rlr
st
ststst
st
dcdc
rlrlrl
λ
λωω
ωω
α
Fall 2005 EE595S Electric Drive Systems 67
15.3 Case Study: Current-Regulated Inverter-Based Seed Control
• Chief Assumption(15.13-1)
• We’ll Use(15.13-2)
• Thus We Obtain
(15.13-3)
∗= ee TT
))(1( 1rmrmspe KT ωωτ −+= ∗∗
τ
τ τ
ωω
JK
JK
JK
rm
rmss
s
++
+=∗ 2
)1(
Fall 2005 EE595S Electric Drive Systems 68
15.3 Case Study: Current-Regulated Inverter-Based Seed Control
• Let Us Place Poles Ats=-5, s=-50
• This YieldsKp=0.257 Nms, τ=0.22 s
• Other Design AspectsUse Sine Triangle Current Regulator From Example 15 B (s=-200, s=-1000)Limit Currents to +/- 3.68 AInject All Current into Q-Axis
Fall 2005 EE595S Electric Drive Systems 69
15.3 Case Study: Current-Regulated Inverter-Based Seed Control
Fall 2005 EE595S Electric Drive Systems 70
15.3 Case Study: Current-Regulated Inverter-Based Seed Control