s.d. sudhoff fall 2005 - purdue university
TRANSCRIPT
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EE595S: Class Lecture NotesChapter 15: Brushless DC Motor Drives
S.D. Sudhoff
Fall 2005
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Fall 2005 EE595S Electric Drive Systems 2
15.1 Architecture
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Fall 2005 EE595S Electric Drive Systems 3
15.2 Topology
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Fall 2005 EE595S Electric Drive Systems 4
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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Fall 2005 EE595S Electric Drive Systems 5
180 VSI Operation
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Fall 2005 EE595S Electric Drive Systems 6
180 VSI Operation
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Fall 2005 EE595S Electric Drive Systems 7
180 VSI Operation
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Fall 2005 EE595S Electric Drive Systems 8
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
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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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Fall 2005 EE595S Electric Drive Systems 10
180 VSI Operation
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Fall 2005 EE595S Electric Drive Systems 11
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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Fall 2005 EE595S Electric Drive Systems 12
Duty Cycle Modulation
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Fall 2005 EE595S Electric Drive Systems 13
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
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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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Fall 2005 EE595S Electric Drive Systems 15
Sine-Triangle Modulation
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Fall 2005 EE595S Electric Drive Systems 16
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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Fall 2005 EE595S Electric Drive Systems 17
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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Fall 2005 EE595S Electric Drive Systems 18
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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Fall 2005 EE595S Electric Drive Systems 19
Modulation Summary
• Note, for space vector modulation(15.3-27))( ** r
dsrqsv jvvangle −=φ
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Fall 2005 EE595S Electric Drive Systems 20
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
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Fall 2005 EE595S Electric Drive Systems 21
Working Out the Details…
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Fall 2005 EE595S Electric Drive Systems 22
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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Fall 2005 EE595S Electric Drive Systems 23
NLAM Startup PredictionFigure 15.6-2
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Fall 2005 EE595S Electric Drive Systems 24
Detailed Startup PredictionFigure 15.6-1
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Fall 2005 EE595S Electric Drive Systems 25
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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Fall 2005 EE595S Electric Drive Systems 26
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
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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
+
−′−=
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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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Fall 2005 EE595S Electric Drive Systems 29
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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Fall 2005 EE595S Electric Drive Systems 30
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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Fall 2005 EE595S Electric Drive Systems 31
Steady-State Performance of VSI Drives
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Fall 2005 EE595S Electric Drive Systems 32
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
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Fall 2005 EE595S Electric Drive Systems 33
Transient Performance of VSI Drives
• In (15.6-1), Note that(15.6-2)xxx ∆+= 0
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Fall 2005 EE595S Electric Drive Systems 34
Predictions of Linearized ModelStartup
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Fall 2005 EE595S Electric Drive Systems 35
Predictions of Detailed ModelStartup
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Fall 2005 EE595S Electric Drive Systems 36
Predictions of Linearized ModelStep Change in Duty Cycle
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Fall 2005 EE595S Electric Drive Systems 37
Predictions of Detailed ModelStep Change in Duty Cycle
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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
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Fall 2005 EE595S Electric Drive Systems 39
Control Law
• Mathematically
(15.8-1)∫ −+−= ∗∗ dtKd rmrmK
rmrm )()( ωωωω τ
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Fall 2005 EE595S Electric Drive Systems 40
Starting Point: Open Loop Frequency Response
• ObserveGain Margin is InfinitePhase Margin is 20o
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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
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Fall 2005 EE595S Electric Drive Systems 42
Compensated Frequency Response
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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)
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Fall 2005 EE595S Electric Drive Systems 44
Compensated Frequency Response
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Fall 2005 EE595S Electric Drive Systems 45
Time-Domain Response
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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
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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
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Fall 2005 EE595S Electric Drive Systems 48
Hysteresis Current-Regulated Drive
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Fall 2005 EE595S Electric Drive Systems 49
Performance of HysteresisControlled Drive
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Fall 2005 EE595S Electric Drive Systems 50
Voltage Source Based Current Regulation
• AdvantagesControlled Switching FrequencyCan Offer Excellent Waveform Quality
• DisadvantageLower Control Bandwidth
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Fall 2005 EE595S Electric Drive Systems 51
Sine Triangle Current Regulator
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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
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Fall 2005 EE595S Electric Drive Systems 53
Example 15B
• Results(15B-4)(15B-5)
sKi /2280Ω=
Ω= 7.10pK
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Fall 2005 EE595S Electric Drive Systems 54
Performance During Step Change In Current
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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−=
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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
=
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Fall 2005 EE595S Electric Drive Systems 57
Response to Step Decrease in DC Voltage
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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
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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+=
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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
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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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Fall 2005 EE595S Electric Drive Systems 62
15.11 Current Command Synthesis(Salient Machines)
• Result
• In this Case, What If Inadequate Voltage
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Fall 2005 EE595S Electric Drive Systems 63
15.11 Current Command Synthesis(Salient Machines)
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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 ′+−++= ∗∗∗∗
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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
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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
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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(
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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
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Fall 2005 EE595S Electric Drive Systems 69
15.3 Case Study: Current-Regulated Inverter-Based Seed Control
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Fall 2005 EE595S Electric Drive Systems 70
15.3 Case Study: Current-Regulated Inverter-Based Seed Control