EE595S: Class Lecture Notes Chapter 14: Induction …sudhoff/ee595s/im_drives.pdf · Chapter 14:...

EE595S: Class Lecture NotesChapter 14: Induction Motor Drives

S.D. Sudhoff

Fall 2005

Fall 2005 EE595S Electric Drive Systems 2

Overview of Strategies

• Volts-Per-Hertz Control• Constant Slip Control• Field-Oriented Control


Volts-Per-Hertz Control

• History

• Advantages

• Disadvantages



• Basic IdeaMachine will operate close to synchronous speed, thus speed controlled with frequencyWe must avoid saturation

• On Avoiding Saturation(14.2-1)(14.2-2)

asassas pirv λ+=

sesV Λ=ω



• Elementary Volts Per Hertz Control



• Example System50 Hp MachineRated for 1800 rpm, 460 V l-l rms, P=4, 60 Hzrs=72.5 mΩ; rr’=41.3 mΩLM=30.1 mH; Lls=Llr=1.32 mH

(14.2-3)⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+=

29.0)(1.0

bmrm

rmbL STTωωω



• Performance Characteristics



• Comments on Steady-State Performance



• Problem 1: Resistance At Low Speed

• Solution: Set Voltage So Slope is Invarient

(4.9-19)

(14.2-4)2,

22,

2,

22,

estssbests

estsseestsbs

Lr

LrVV

ω

ω

+

+=

222

22

22

)()(

~2

3

rrsssrbe

rrssMbe

rs

asrbM

be

e

XsrXrXXXsrr

VsrXP

T

′+′⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡′−⎟⎟

⎠

⎞⎜⎜⎝

⎛+′

′⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

=

ωω

ωω

ωωω



• Further Improvement – Current FeedbackLet’s approximate torque as

• (14.2-5)• (14.2-6)

Using (14.2-4) we have

• (14.2-7)

• This is not a function of synchronous speed

)( retve KT ωω −=

erre

tvTK

ωωω =∂∂

−=

)(2

3

2222

22

ssbsr

brMtv

Lrr

VrLP

Kω+′

′⎟⎠⎞

⎜⎝⎛

=



Next, consider torque. We have• (14.2-8)

For steady-state conditions• (14.2-9)

• (14.2-10)

Thus• (14.2-11)

Where• (14.2-12)

)(22

3 eds

eqs

eqs

edse iiPT λλ −=

e

eqss

eqse

dsirv

ωλ

−=

e

edss

edse

qsirv

ωλ

−−=

)2(122

3 2*ss

eqs

eqs

ee IrivPT −=

ω

222

1 eds

eqss iiI +=



Equating (14.2-7) and (14.2-11)

• (14.2-13)

Or• (14.2-14)

Where• (14.2-15)

• (14.2-16)

2/)2(3 2*2**

tvsseqs

eqsrr

eKIrivP −++

=ωω

ω

2),0max( 2**

corrrre

Χ++=

ωωω

corrcorr sH LPF χ)(=Χ

tvsseqs

eqscorr KIrivP /)2(3 2* −=χ



• Performance of Compensated Volts-Per-Hertz Control



• Start-Up Performance of Compensated Volts Per Hertz Control


Constant Slip Control

• Basic IdeaDefinition of Slip Frequency

(14.3-1)Our Strategy

res ωωω −=


Derivation of Constant Slip Control

• Consider Our Expression for Electromagnetic Torque

• (14.3-2)

• Solving for the Current Yields

• (14.3-3)

• The Plan

22

22

)()(2

3

rrsr

rsMse

Lr

rILP

T′+′

′⎟⎠⎞

⎜⎝⎛

=ω

ω

estrrestMs

estrrsestres

rLP

LrTI

,2

,

2,

2,

*

||3

))((||2

′

′+′=

ω

ω


Avoiding Saturation

• Before Proceeding …

• Consider the Steady-State Equivalent Circuit

(14.3-4)(14.3-5)

Thus(14.3-7)

)~~(~~arasMarlrar IILIL ′++′=λ

srLjLjII

rrreMe

asar /~~

′+′−=′

ωω

222'

rrrs

rMsr

rL

rLI′+′

′=

ωλ


Avoiding Saturation

Taking the Magnitude(14.3-7)

Combing (14.3-7) and (14.3-2)

(14.3-8)Thus

(14.3-9)So For Large Torque Commands

(14.3-10)

222'

rrrs

rMsr

rL

rLI′+′

′=

ωλ

rrs

rePT ′=

2'

23 λω

estr

rsetsthreshe r

PT,

2max,,

, 23

′=

λω

2max,

,*

3

2

r

estres

P

rT

λω

′=


Constant Slip Control


Selection of Slip Frequency

• Control for Maximum Torque Per AmpFrom (14.3-1)

(14.3-11)

Optimizing with Respect to Slip Frequency

(14.3-12)

2,

2

2,

2 )()(2

3

rsetsr

rMsets

s

eLr

rLP

IT

′+′

′⎟⎠⎞

⎜⎝⎛

=ω

ω

estrr

estrsets L

r

,

,, ′

′=ω


Maximum Torque Per Amp Control


Maximum Efficiency Control

• To Optimize Efficiency Note That(14.3-13)

So(14.3-14)

Comparison to (14.3-14) to (14.3-2)(14.3-15)

Now (14.3-16)

)~Re(3 assin VIP =

22

222

)(33

rrsr

rsMesssin

LrrLIIrP

′+′

′+=

ω

ωω

eessin TP

IrP ω23 2 +=

erout TP

P ω2=



So(14.3-17)

Finally

(14.3-18)

Minimizing Yields(14.3-19)

sessoutin TP

IrPP ω23 2 +=−

⎥⎥⎦

⎤

⎢⎢⎣

⎡+

′+

′= s

mr

rrss

ms

sreloss

LrLr

LrrT

PP ωω

ω 2

2

22

1

1

,

,2

,

2,,

,,

+′′

′

′=

estr

ests

estrr

estmestrr

estrsets

rr

L

LLr

ω


Comparison of Constant Slip Controls

• Maximum Efficiency

• Maximum Torque Per Amp


Speed Control with Constant Slip Control


Transient Performance of Constant Slip Controlled Drive


14.4 Field-Oriented Control

• Objective

• Challenges


The Basic Idea

θsin2BiNLrTe −=


Torque Production in Induction Motor

• Now(14.4-2)

(14.4-3)

)(22

3qrdrdrqre iiPT ′′−′′= λλ

θλ sin||||22

3qdrqdre iPT ′′−=


The Plan

• Objective

• ObservationIn Steady-State, This is Always True

• (14.4-4)

• (14.4-5)

• (14.4-6)

Thus

edrre

reqr ri λωω )(1

−′

−=

eqrre

redr ri λωω )(1

−′

=

edr

edr

eqr

eqr

eqdr

eqdr iii ′′+′′=′⋅′ λλλ


The Plan (Continued)

• But.. We need flux and current to be perpendicular all the time. To do this ..

(14.4-7)(14.4-8)

• Achieving (14.4-7)By choice of θe

0=′eqrλ

0=′edri


The Plan (Continued)

• Achieving (14.4-8)All we need to do is to keep d-axis stator current constant. Proof:

• Consider(14.4-9)

• The q-axis rotor flux linkage is zero (by choice of reference frame)

(14.4-10)• Substitution of d-axis rotor flux linkage equation

into (14.4-10)(14.4-11)

• Thus the d-axis rotor current goes to zero

edr

eqrre

edrr pir λλωω ′+′−+′′= )(0

edr

edrr pir λ′+′′=0

eds

rrMe

drrrre

dr piLLi

Lrip −′′

−=′


Some Observations

• Since the d-axis rotor current is zero

(14.4-12)

(14.4-13)

• Combining these results with our flux linkage equation

(14.4-14)

edsss

eds iL=λ

edsM

edr iL=′λ

eqr

edre iPT ′′−= λ

223


Some Observations

• Since the q-axis rotor flux linkage is zero

(14.4-15)

• Thus

(14.4-16)

eqs

rrMe

qr iLLi −=′

eqs

edrrr

Me i

LLPT λ′=

223


Generic Rotor Flux-Oriented Control

Note: θe must be selected so as to keep q-axis rotor flux equal to zero.


Some “Minor” Details

• Determination of position of synchronous reference frame

• Determination of the rotor flux


Aside Types of Field Oriented Control

• OrientationsStatorMagnetizing (Air-Gap)Rotor

• MethodsDirectIndirect

• ModificationsRobust


Direct Field Oriented Control

• Basic IdeaMeasure the flux directly (more or less)

• Consider

(14.5-1)

• We need to set q-axis rotor flux to zero. Thus

(14.5-2)

⎥⎥⎦

⎤

⎢⎢⎣

⎡

′

′⎥⎦

⎤⎢⎣

⎡ −=

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

′

′sdr

sqr

ee

eee

dr

eqr

λ

λθθθθ

λ

λcossinsincos

2)( πλλθ +′−′= s

drsqre jangle


Direct Field-Oriented Control

• As a Side Effect

(14.5-3)

• The problem: We can measure the rotor flux linkages. We can measure the stator flux linkages.

22 )()( sdr

sqr

edr λλλ ′+′=′


Estimation of Rotor Flux Linkages

• Consider(14.5-4)

• Thus(14.5-5)

• Recall(14.5-6)

)( sqr

sqsM

sqm iiL ′+=′λ

M

sqsM

sqms

qr LiL

i−′

=′λ

)( sqr

sqsM

sqrlr

sqr iiLiL ′++′=′λ


Estimation of Rotor Flux Linkages

• So(14.5-7)

(14.5-8)

sqslr

sqm

Mrrs

qr iLLL ′−′

=′ λλ

sdslr

sdmM

rrsdr iL

LL ′−′

=′ λλ


Rotor Flux Estimator


Direct Rotor Field-Oriented Control


Robust Direct Field-Oriented Control

• Sensitivity of Rotor Flux Estimator(14.6-1)

• Sensitivity of Q-Axis Current Calculation

(14.6-2)

sqdsestlr

smqdestM

estrreqdr iL

LL

,,,

,ˆ ′−′

=′ λλ

edrestrr

estMee

qs

LLP

Tiλ′

=ˆ

223

,

,

**



• Sensitivity of the D-Axis Current Injection

(14.6-3)estM

edre

ds Li

,

** λ=


Flux Control Loop

• Since we have a flux estimator, consider

• It can be shown that

(14.6-5)1

1,*

+

+=

′

′

sL

Ls

M

estMedr

edr

λ

λ

τ

τ

λ

λ


Torque Control Loop

• Torque EstimationRecall

(14.6-6)Now

(14.6-7)So

(14.6-8)Which Suggests

(14.6-9)

)(22

3 sds

sqs

sqs

sdse iiPT λλ −=

sqdm

sqdsls

sqds iL λλ +=

)(22

3 sds

sqm

sqs

sdme iiPT λλ −=

)(22

3ˆ sds

sqm

sqs

sdme iiPT λλ −=


Torque Control Loop

• Now that we can estimate torque, how about


Torque Control Loop

• Analysis of Torque Control LoopDefine

(14.6-10)(14.6-12)

Now, assuming the control works(14.6-11)

This, assumption, coupled with our torque control loop yields

(14.6-13)

*,

,, 22

3 edrestrr

estMestt L

LPK λ′′

=

edrrr

Mt L

LPK λ′′

=22

3

*eqste iKT =

1

1,*

+

+=

sK

Ks

TT

t

esttt

t

e

e

τ

τ


Performance of Robust Direct Field Oriented Control


Performance of Constant Slip Drive


Performance of Volts-Per-Hertz Drive


Indirect Field Oriented Control

• Consider (14.7-1)

Since the q-axis rotor flux is zero

(14.7-2)

Which may be expressed

(14.7-3)

eqr

edrre

eqrrr pir λλωω ′+′−+′′= )(0

edr

eqr

rrei

rλ

ωω′

′′−=

ede

eqs

rrr

rei

iLr′′

+= ωω



• ThoughtWe know what the electrical frequency will be if we have field orientation (i.e. field orientation causes (14.7-1))Does it work backward ? Will using (14.7-1) cause field-orientation ?The answer: Yes !


Indirect Field Orientation

• ProofConsider the rotor voltage equations

(14.7-5)(14.7-6)

Now substitute in our expression for electrical frequency

(14.7-7)

(14.7-8)

eqr

edrre

eqrr pir λλωω ′+′−+′′= )(0

edr

eqrre

edrr pir λλωω ′+′−−′′= )(0

eqr

edre

de

eqs

rrre

qrr pi

iLrir λλ ′+′′′

+′′= *

*0

edr

eqre

de

eqs

rrre

drr pi

iLrir λλ ′+′′′

−′′= *

*0



• Continuing On …Now put in terms of q-axis rotor flux and d-axis rotor current (and stator currents)This yields

(14.7-9)

(14.7-10)

[ ] eqr

edsM

edrrre

ds

eqs

rr

r

rr

eqsM

eqr

r piLiLi

iLr

LiL

r λλ

′++′′′′

+⎥⎥⎦

⎤

⎢⎢⎣

⎡

′

−′′= *

*

**0

[ ]**

*0 e

dsMe

drrreqre

ds

eqs

rr

redrr iLiLp

i

iLrir +′′+′′′

−′′= λ



• Now, keeping the d-axis stator current fixed we have

(14.7-11)

(14.7-12)

• Comments

edre

ds

eqs

reqr

rrre

qr ii

ir

Lrp ′′−′′′

−=′*

*λλ

eqre

ds

eqs

rr

redrrr

redr i

i

Lri

Lrip λ′

′

′+′

′′

−=′*

*

2)(



• Thus, our control becomes



• Advantages of Indirect Field Oriented Control over Direct Field Oriented Control

• Disadvantages of Indirect Field Oriented Control over Direct Field Oriented Control


Performance of DetunedIndirect Field Oriented Control

• Overestimate magnetizing inductance by 25% (we started to saturate machine)

• Underestimate rotor resistance by 25% (rotor resistance increased with temperature)


Performance of Detuned Field Oriented Control

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