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

15.2 Topology

15.3 Equivalence of VSI Schemesto Idealized Source

• 180 VSI• Duty Cycle Modulation• Sine-Triangle Modulation• Sine-Triangle Modulation (with 3rd)• Space-Vector Modulation

180 VSI Operation

• Comparing the two figures(15.3-1)

• Now recall(15.3-2)

(15.3-3)

hrc φθθ +=

dccqs vv π

0=cdsv

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

hv φφ =

180 VSI Operation

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

0=cdsv

hdcrqs vdv φπ cos2=

hdcrds vdv φπ sin2−=

dcs vdvπ2

hv φφ =

Duty Cycle Modulation

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 +=

⎪⎩

⎪⎨⎧

0=cdsv

( ) 11arccos21)( 4121

21 >⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−+−= d

dddf d π

• Converting to the rotor reference frame

(15.3-17)

• (15.3-18)

⎪⎩

⎪⎨⎧

1cos)(

vdcrqs φ

⎪⎩

⎪⎨⎧

≤−=

1sin)(

vdcrds φ

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

Space Vector Modulation

• In this case we have

(15.3-21)

(15.3-22)

(15.3-23)

⎪⎪⎩

⎪⎪⎨

dcspkspk

dcspkrqs

rqs vv

⎪⎪⎩

⎪⎪⎨

dcspkspk

dcspkrds

rds vv

( ) ( )2*2** rds

rqsspk vvv +=

Modulation Summary

• For all VSI strategies, we have(15.3-24)(15.3-25)

• Where

(15.3-26)

vdcrqs mvv φcos=

vdcrds mvv φsin−=

( )( )⎪

⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪⎪

≥−

≤−

3/vvvv

dc*spk

modulationvectorpace3

modulationvectorspace

)3/2( modulation harmonic-third

1) (modulation triangle-sine)(

1)( modulation triangle-sine

modulation cycle-duty

modulation source- voltage180

Modulation Summary

• Note, for space vector modulation(15.3-27))( ** r

dsrqsv jvvangle −=φ

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

⎪⎩

⎪⎨⎧

rectifierphase single rectifierphase three

⎪⎩

⎪⎨⎧

ceur L

rectifierphase single rectifierphase three2

⎩⎨⎧

Working Out the Details…

Final Average Value Model

• Nonlinear Average Value Model (NLAM)

(15.4-21)( )

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢⎢

−−+−

−−

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

′−

−−

−−−

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢⎢

))((sin

)sin(coscos

000000000000

00000000000000000000

TiiLLimv

iivivi

Liivivi

φφα

NLAM Startup PredictionFigure 15.6-2

Detailed Startup PredictionFigure 15.6-1

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

rqsdc iimi φφ −=

)sincos(0 23

rqsstr IImII φφ −−−=

)sincos(0 23

rqsr IImI φφ −−=

• 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

rdssvdc ILIrmV ωφ +−−= sin0

• Solving (15.5-5) and (15.5-6) for the Currents

(15.5-7)

(15.5-8)

vdcdrmrvdcsrqs

LLrmVLmVrI 22

sin)cos(ω

φωλωφ+

+′−=

vdcsmrvdcqrrds LLr

mVrmVLI 22

sin)cos(

φλωφω

−′−=

• 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

−+++

−′++=

• 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(

vqrvsmrIdc LLrm

LrV dc φω

φωφλω

−′==

• 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

15.6 Transient Performance of VSI Drives

• Linearizing (15.4-21) We Have

(15.6-1)

( ) ( )

⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢

∆∆∆∆∆

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

−−

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢⎢

∆∆

∆∆∆∆∆

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

−−+′

−+−

−−−−

−−

=∆∆∆∆∆∆∆

rqsqdJ

PrdsqdmJ

LstLstr

rqsststdcr

iivivi

iLLiLL

iivivip

stdcdcdcdc

φφφφ

0cossin00

0sincos000000000000000

0)())((0000

00sin0

00cos0

0000000000

0sincos00

)cos(sin23)sin(cos

000000000

Transient Performance of VSI Drives

• In (15.6-1), Note that(15.6-2)xxx ∆+= 0

Predictions of Linearized ModelStartup

Predictions of Detailed ModelStartup

Predictions of Linearized ModelStep Change in Duty Cycle

Predictions of Detailed ModelStep Change in Duty Cycle

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

Control Law

• Mathematically

(15.8-1)∫ −+−= ∗∗ dtKd rmrmK

rmrm )()( ωωωω τ

Starting Point: Open Loop Frequency Response

• ObserveGain Margin is InfinitePhase Margin is 20o

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

Compensated Frequency Response

Selection of K

• Select K to Achieve Phase Margin of 60o

• Thus, We Choose K=0.25 (-12 dB)

Compensated Frequency Response

Time-Domain Response

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

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

Hysteresis Current-Regulated Drive

Performance of HysteresisControlled Drive

Voltage Source Based Current Regulation

• AdvantagesControlled Switching FrequencyCan Offer Excellent Waveform Quality

• DisadvantageLower Control Bandwidth

Sine Triangle Current Regulator

Example 15B

• Suppose We Utilize

(15.B-1)

(15.B-2)

• Place Poles At s=-200, s=-1000

)()( rqs

rdsssr

rqs ii

sKKiLv −⎟

⎠⎞

⎜⎝⎛ ++′+= ∗∗ λω

)( rds

rqsssr

rds ii

sKKiLv −⎟

⎠⎞

⎜⎝⎛ ++−= ∗∗ ω

Example 15B

• Results(15B-4)(15B-5)

sKi /2280Ω=

Ω= 7.10pK

Performance During Step Change In Current

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

rqs iLirv λωω ′++= **

** rqsqr

rds iLirv ω−=

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

dsrqss vvv +=

22 )()(2

1 ∗∗∗∗ −+′++= rqsqr

rdssrm

rqsss iLiriLirv ωωλω

dcs vv6

dcs vvπ2

Response to Step Decrease in DC Voltage

15.11 Current Command Synthesis(Non-Salient Machines)

• Recall (15.11-1)

• Thus(15.11-2)(15.11-3)

• Advantages

• But..

Pe iT λ′= 22

∗′

∗ = ePrqs Ti

0=∗rdsi

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

22222 )(2

izrvzLi

rqsmrssrssmr

∗∗ +′−+′−=

λωωλ

222ssrs Lrz ω+=

15.11 Current Command Synthesis(Salient Machines)

• Recall(15.11-7)

• Thus(15.11-8)

• Also Recall(15.11-9)

))((223 r

dsrqsqd

Pe iiLLiT −+′= λ

rqsqdP

erds LLiLL

Ti−′

−−

= ∗∗ λ1

22 )()(2

1 ∗∗ += rds

rqss iii

• Strategy: Minimize Fundamental Component of Current

• Why?

• Result

• In this Case, What If Inadequate Voltage

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 ii =

∗= rds

rds ii

rmrdsdr

rqs iLirv ωλω ′++= ∗∗∗

∗∗ −= rqsqr

rds iLirv ω

])()([ 223 r

qsmrrds

rqsqdr

rqss iiiLLiirP λωω ′+−++= ∗∗∗∗

• Since(15.12-6)

• We Have(15.12-7)

• Again Assuming Our Currents Are Equal to Commanded Values

(15.12-8)

dcdc V

])()([ 2123 r

qsmrrds

rqsqdr

rqssVdc iiiLLiiri

dcλωω ′+−++= ∗∗∗∗

))((223 ∗∗∗ −+′= r

dsrqsqd

Pe iiLLiT λ

• Resulting State Model(15.12-9)

⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢

−−+′

′+−++−+

⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢⎢⎢

−−

⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢

∗∗∗

∗∗∗∗∗

]))(([00

])()([

rqsqdr

rqssVC

TiiLLi

iiiLLiir

ststst

rlrlrl

λωω

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 −+= ∗∗

+=∗ 2

• 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

S.D. Sudhoff Fall 2005 - Purdue University

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