EE631 – Spring 2005 1

ECE631/EE631QLecture 16: Average Value Model

S.D.Sudhoff

Purdue University

EE631 – Spring 2005 2

IM Overview

ini

invinr

inC

aLi

bLi

cLi

Lacr

Lacr

Lacr

acL

acL

acL

Cacr

Cacr

Cacr

acC

acC

acC

aOi

bOi

abOv

*eqLi

*edLi

e

*eqOv*edOv

e

aLi

bLi

VoltageController

CurrentController Voltage

ControllerVoltage

Controller

cbOv

EE631 – Spring 2005 3

Objectives For Today

• Develop an Average Value Model for the 3-Phase Bridge Inverter

• Features– Include conduction losses– Ignore switching losses

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Semiconductor Model

• Transistor

• Diode

irvv swsw

irvv dd

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AVM – DC Currrent

• Consider the current reference frame

cicqiai ii cos

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AVM – DC Current

• Sine-Triangle Modulation with 3rd Harmonic

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AVM – DC Current

• We have

)1(21

asag dd

))(3cos()cos( 3 cicias ddd

EE631 – Spring 2005 8

AVM – DC Current• We have

agaidca dii ,ˆ

2/3

2/,, )(ˆ

2

1

cicidcadca dii

EE631 – Spring 2005 9

AVM – DC Current

EE631 – Spring 2005 10

AVM – DC Current

• Thus we have for the a-phase

• The total dc current must be given by

)cos(4

1, ci

qidca idi

)cos(4

3 ciqidc idi

EE631 – Spring 2005 11

AVM - Voltage• The fast average voltages may be expressed

0ˆ

0ˆˆ

aiag

aiagag iv

ivv

)1)(()(ˆ agaiddagaiswswdcag dirvdirvvv

)1)(()(ˆ agaiswswagaidddcag dirvdirvvv

EE631 – Spring 2005 12

AVM - Voltage

• The fundamental component of the line-to-ground voltage may be expressed

cicifundag bav sincosˆ 11

ciciagciciag dvdva

2/3

2/

2/

2/1 )cos(ˆ

1)cos(ˆ

1

ciciagciciag dvdvb

2/3

2/

2/

2/1 )sin(ˆ

1)sin(ˆ

1

EE631 – Spring 2005 13

AVM - Voltage

• After substantial manipulation

ciqcdswdsw

swddswdc

irrddrr

vvdvvva

)3cos()cos(3

4)(

2

1

)(2

cos)(2

1

351

1

ciqc

dsw

swddc

iddrr

dvvvb

)3sin(5

3)sin(

3

)(2

)sin()(2

1

3

1

EE631 – Spring 2005 14

AVM - Voltage• After a little more work, we have

• To show this

1av ciqi

1bv cidi

EE631 – Spring 2005 15

AVM - Voltage

EE631 – Spring 2005 16

AVM – Modulation Signal

• Problem: we need to know d, d3, • Starting point: modulation indices

dc

qq v

vm

**

dc

dd v

vm

**

EE631 – Spring 2005 17

AVM – Modulation Signal

• We will need the modulation signal in the current reference frame

*

*

*

*

)cos()sin

)sin()cos(

d

q

cici

cicicid

ciq

m

m

m

m

EE631 – Spring 2005 18

AVM – Modulation Signal

• It can be shown that

• To show this

)angle( diqici jii

EE631 – Spring 2005 19

AVM – Modulation Signal

EE631 – Spring 2005 20

AVM – Modulation Signal

EE631 – Spring 2005 21

AVM – Modulation Signal

• Lets also define a voltage reference frame

• In this frame)( **dqcv jmmangle

2*2**dq

cvq mmm

0* cvdm

)3cos()cos( *3

*cvcvas ddd

EE631 – Spring 2005 22

AVM – Modulation Signal

• To show this

EE631 – Spring 2005 23

AVM – Modulation Signal• Since we have

• We therefore have

)angle( diqici jii

)(angle)angle( **dqdiqi jmmjii

)( **dqcv jmmangle

))(3cos()cos( 3 cicias ddd

)3cos()cos( *3

*cvcvas ddd

EE631 – Spring 2005 24

AVM Modulation Signal

• Desired magnitudes

• For unsaturated operation

** 2 cvqmd

**3 6

1dd

3

2* d

*dd

*33 dd

EE631 – Spring 2005 25

AVM - Modulation

• Analysis plan for saturated operation

)3cos()cos( 3, cvcvaprxa ddd

2/

02

,

3

)())(,1,1bound(

),( cvcvaprxacva ddd

ddf

bxb

bxax

axa

xba ),,bound(

EE631 – Spring 2005 26

AVM Modulation

• Results

1*

1*3

**311

*

2)4sin(

2

1

))(2sin(1

sin4

cvcv

cvcv

dd

dddd

111

*3

1*3

*1

*3

*

1*3

*33

sin)2cos(845cos2

)3cos()23(cos)29(

3

1

2

cvcvcv

cvcv

cv

d

dddd

ddd

1)3cos()cos( 131 cvcv dd

EE631 – Spring 2005 27

AVM Modulation

• A subtle point

• This occurs when 5

6

3

2 d

EE631 – Spring 2005 28

AVM – Hysteresis Modulator