MAGNETICALLY COUPLED NETWORKSLEARNING GOALS

Mutual Inductance Behavior of inductors sharing a common magnetic field

The ideal transformer Device modeling components used to change voltage and/or current levels

BASIC CONCEPTS – A REVIEW

Magnetic field Total magnetic flux linked by N-turn coil

Ampere’s Law(linear model)

Faraday’sInduction Law

Assumes constant L and linear models!

Ideal Inductor

MUTUAL INDUCTANCEOverview of Induction Laws

Magneticflux

webers) linkageflux Total

( N

Li

If linkage is created by a current flowing through the coils…

(Ampere’s Law)

The voltage created at the terminals of the components is

dtdiLv (Faraday’s Induction Law)

Induced linkson secondcoil

2( )

One has the effect of mutual inductance

TWO-COIL SYSTEM (both currents contribute to flux)

Self-induced Mutual-induced

Linear model simplifyingnotation

THE ‘DOT’ CONVENTIONCOUPLED COILS WITH DIFFERENT WINDING CONFIGURATION

Dots mark reference polarity for voltages induced by each flux

THE DOT CONVENTION REVIEW

Currents and voltages followpassive sign convention

Flux 2 inducedvoltage has + at dot

)()()(

)()()(

22

12

2111

tdtdiLt

dtdiMtv

tdtdiMt

dtdiLtv

LEARNING EXAMPLE

)(1 ti )(2 ti

))(( 2 tv

For other cases change polarities orcurrent directions to convert to thisbasic case

dtdiM

dtdiLtv 21

11 )(

dtdiL

dtdiMtv 2

21

2 )(

dtdiL

dtdiMv

dtdiM

dtdiLv

22

12

2111

LEARNING EXAMPLE

Mesh 1

LEARNING EXAMPLE - CONTINUED

Mesh 2 Voltage Terms

1i

1v

dtdiL

dtdiMv

dtdiM

dtdiLv

22

12

2111

)()()(

)()()(

22

12

2111

tdtdiLt

dtdiMtv

tdtdiMt

dtdiLtv

Equivalent to a negative mutualinductance

2i

2vdtdiM

dtdiLv 21

11

dtdiL

dtdiMv 2

21

2

More on the dot convention

LEARNING EXTENSION

21 1( ) ( ) ( )didiv t L t M t

dt dt

1 22 2( ) ( ) ( )di div t M t L t

dt dt

21 1( ) ( ) ( )didiv t L t M t

dt dt

1 22 2( ) ( ) ( )di div t M t L t

dt dt

Convert tobasic case

)(),( 21 tvtvfor equations the Write

2212

2111

ILjMIjVMIjILjV

Phasor model for mutually

coupled linear inductorsAssuming complex exponential sources

PHASORS AND MUTUAL INDUCTANCE

)()()(

)()()(

22

12

2111

tdtdiLt

dtdiMtv

tdtdiMt

dtdiLtv

LEARNING EXAMPLEThe coupled inductors can be connected in four different ways.Find the model for each case

CASE I

1V 2V ILjMIjVMIjILjV

22

11

Currents into dots

CASE 2

1V 2V

Currents into dotsI I

I I 21 VVV

21 VVV

1 1

2 2

V j L I j MIV j MI j L I

1 2( 2 ) eqV j L L M I j L I

ILMLjV )2( 21

eqL

M

Leq of valuethe on

constraint physical a imposes 0

CASE 3 Currents into dots

1I 2I

V V

21 III

221

211

ILjMIjVMIjILjV

12 III

)()(

121

111

IILjMIjVIIMjILjV

ILjIMLjVMIjIMLjV

212

11

)()(

)/()/(

1

2

MLML

IMLLMLMjVMLL )()()2( 12221 I

MLLMLLjV221

221

CASE 4Currents into dots

1I 2I

V )( V 21 III

221

211

ILjMIjVMIjILjV

21 2

1 2 2L L MV j IL L M

LEARNING EXAMPLE0V VOLTAGETHE FIND

1V

2V

2I

1123024 VI :KVL022 222 IIjV- :KVL

)(62)(24

212

211

IjIjVIjIjV

CIRCUIT INDUCTANCEMUTUAL 20 2IV

SV

21

21

)622(202)42(

IjjIjIjIjVS

1I

42/2/j

j

22)42(42 IjVj S

1682

2 jVjI S

jj

jVS816

2

jVIV S

242 20

57.2647.4

3024 42.337.5

1. Coupled inductors. Define theirvoltages and currents

2. Write loop equationsin terms of coupledinductor voltages

3. Write equations forcoupled inductors

4. Replace into loop equationsand do the algebra

LEARNING EXAMPLEWrite the mesh equations

1V

21 II

2V

32 II

1. Define variables for coupled inductors2. Write loop equations in terms of coupled inductor voltages

1

21111 Cj

IIVIRV

0)(1

123232221

CjIIIIRVIRV

0)( 233342

32 IIRIR

CjIV

)()()()(

322212

322111

IILjIIMjVIIMjIILjV

3. Write equations for coupled inductors

4. Replace into loop equations and rearrange terms

321

1

11

11

1

1

MIjICj

MjLj

ICj

LjRV

332

21

3222

11

1

1

10

IRLjMj

ICj

RLjMjRMjLj

ICj

MjLj

3342

2

2321

1

0

IRRCj

Lj

IRMjLjMIj

LEARNING EXAMPLE

)(13 jZS )(11 jZL

)(21 jLj )(22 jLj)(1 jMj

DETERMINE IMPEDANCE SEEN BY THE SOURCE1IVZ S

i

1I 2I

1V

2V

SS VVIZ 11

1. Variables for coupled inductors

2. Loop equations in terms of coupled inductors voltages

022 IZV L

3. Equations for coupled inductors)( 2111 IMjILjV )( 2212 ILjMIjV

4. Replace and do the algebra

0)()(

)(

221

211

ILjZIMjVIMjILjZ

L

SS

MjLjZL

/)/( 2

SL

LS

VLjZIMjLjZLjZ

)()())((

2

12

21

2

2

11

)()(LjZ

MjLjZIVZ

LS

Si

11)1(332

jjjZ i

j

j

1133

jj

11

)(5.25.32133

jjjZ i

)(54.3530.4 iZ

THE IDEAL TRANSFORMER

Insures that ‘no magnetic fluxgoes astray’

11 N 22 N

2

1

2

1

22

11

)()(

)()(

NN

vv

tdtdNtv

tdtdNtv

First ideal transformerequation

0)()()()( 2211 titvtitv Ideal transformer is lossless

1

2

2

1

NN

ii

Second ideal transformerequations

1

2

2

1

2

1

2

1 ;NN

ii

NN

vv

Circuit Representations

REFLECTING IMPEDANCES

dots)at signs (both 2

1

2

1

NN

VV

r)transforme leaving I(Current 21

2

2

1

NN

II

Law) s(Ohm' 22 IZV L

2

11

1

21 N

NIZNNV L 1

2

2

11 IZ

NNV L

LZNNZ

IV

2

2

11

1

1

sideprimary the into reflected , impedance, LZZ 1

For future reference

2*22

*

1

22

2

12

*111 SIV

NNI

NNVIVS

ratio turns 1

2

NNn

21

21

21

21

SSnZZ

nIInVV

L

Phasor equations for ideal transformer

LEARNING EXAMPLEDetermine all indicated voltages and currents25.04/1 n

Strategy: reflect impedance into theprimary side and make transformer“transparent to user.”

21 nZZ L

LZ

16321 jZ

5.1333.25.1342.51

012012500120

1 jI

012021

1111 ZZ

ZIZV

1205.1342.51

16320120

5.1333.2)1632(

21

1

11

j

ZZZ

jIZ

07.1336.835.1333.257.2678.351V

dot) into(current 112 4I

nII

dot) to opposite is ( 112 25.0 VnVV

CAREFUL WITH POLARITIES ANDCURRENT DIRECTIONS!

USING THEVENIN’S THEOREM TO SIMPLIFY CIRCUITS WITH IDEAL TRANSFORMERS

Replace this circuit with its Theveninequivalent

00

121

2

InII

I11 SVV

112

11SOC

S nVVnVVVV

To determine the Thevenin impedance...

THZReflect impedance intosecondary

12ZnZTH

Equivalent circuit with transformer“made transparent.”

One can also determine the Theveninequivalent at 1 - 1’

USING THEVENIN’S THEOREM: REFLECTING INTO THE PRIMARY

Find the Thevenin equivalent ofthis part

00 21 II and circuit open InnVV S

OC2

Thevenin impedance will be the thesecondary mpedance reflected intothe primary circuit

22

nZZTH Equivalent circuit reflecting

into primary

Equivalent circuit reflecting into secondary

LEARNING EXAMPLEDraw the two equivalent circuits

2n

Equivalent circuit reflecting into secondary

Equivalent circuit reflectinginto primary

LEARNING EXAMPLE oV Find

secondary intoreflect tobetter is compute To oV

But before doing that it is better to simplify the primary using Thevenin’s Theorem

Thevenin equivalent of this part

dV

904dOC VV

024444jjVd

)4||4(2 jZTH

14.42 33.69 ( )OCV V

)(24 jZTH441688

44162

jjj

jjZTH

248

11

162 j

jj

jjZTH

j

1

9024

This equivalent circuit is now transferred tothe secondary

LEARNING EXAMPLE (continued…)

Thevenin equivalent of primary side

2n

69.3384.28520

2j

Vo

04.1462.2069.3384.282

Equivalent circuit reflecting into secondary

Circuit with primary transferred to secondary

LEARNING EXAMPLE 2121 ,,, VVIIFind

Nothing can be transferred. Use transformer equations and circuitanalysis tools

21

21

nIInVV

Phasor equations for ideal transformer

022

0101

211 IVVV :1 Node @

022 2212

IjVVV :2 @Node

4 equations in 4 unknowns!

2n21

12

221

121

22

02)1(01022

IIVV

IVjVIVV

051I

05.22I

05)2)(1( 11 VjV

43.6324.205

2105

1 jV 43.635

43.63522V