MAGNETICALLY COUPLED NETWORKS

LEARNING 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 INDUCTANCE

Overview 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

dt

diLv (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’ CONVENTION

COUPLED 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

tdt

diLt

dt

diMtv

tdt

diMt

dt

diLtv

LEARNING EXAMPLE

)(1 ti )(2 ti

))(( 2 tv

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

dt

diM

dt

diLtv 21

11 )(

dt

diL

dt

diMtv 2

21

2 )(

dt

diL

dt

diMv

dt

diM

dt

diLv

22

12

2111

LEARNING EXAMPLE

Mesh 1

LEARNING EXAMPLE - CONTINUED

Mesh 2 Voltage Terms

2212

2111

ILjMIjV

MIjILjV

Phasor model for mutually

coupled linear inductorsAssuming complex exponential sources

PHASORS AND MUTUAL INDUCTANCE

)()()(

)()()(

22

12

2111

tdt

diLt

dt

diMtv

tdt

diMt

dt

diLtv

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

CASE I

1V 2VILjMIjV

MIjILjV

22

11

Currents into dots

CASE 2

1V 2V

Currents into dots

I 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 3Currents into dots

1I 2I

V V

21 III

221

211

ILjMIjV

MIjILjV

12 III

)(

)(

121

111

IILjMIjV

IIMjILjV

ILjIMLjV

MIjIMLjV

212

11

)(

)(

)/(

)/(

1

2

ML

ML

IMLLMLMjVMLL )()()2( 12221 I

MLL

MLLjV

221

221

CASE 4Currents into dots

1I 2I

V )( V 21 III

221

211

ILjMIjV

MIjILjV

2

1 2

1 2 2

L L MV j I

L L M

LEARNING EXAMPLE0V VOLTAGETHE FIND

1V

2V

2I

1123024 VI :KVL

022 222 IIjV- :KVL

)(62

)(24

212

211

IjIjV

IjIjV

CIRCUIT INDUCTANCEMUTUAL

20 2IV

SV

21

21

)622(20

2)42(

IjjIj

IjIjVS

1I

42/

2/

j

j

22)42(42 IjVj S

168

22 j

VjI S

j

j

j

VS816

2

j

VIV 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 inductors

2. Write loop equations in terms of coupled inductor voltages

1

21111 Cj

IIVIRV

0)(1

123232221

Cj

IIIIRVIRV

0)( 233342

32 IIRIR

Cj

IV

)()(

)()(

322212

322111

IILjIIMjV

IIMjIILjV

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 SOURCE1I

VZ Si

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

ILjZIMj

VIMjILjZ

L

SS

Mj

LjZL

/

)/( 2

SL

LS

VLjZ

IMjLjZLjZ

)(

)())((

2

12

21

2

2

11

)()(

LjZ

MjLjZ

I

VZ

LS

Si

11

)1(33

2

j

jjZ i

j

j

1

133

j

j

1

1

)(5.25.32

133

j

jjZ i

)(54.3530.4 iZ

THE IDEAL TRANSFORMER

Insures that ‘no magnetic fluxgoes astray’

11 N 22 N

2

1

2

1

22

11

)()(

)()(

N

N

v

v

tdt

dNtv

tdt

dNtv

First ideal transformerequation

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

1

2

2

1

N

N

i

i Second ideal transformer

equations

1

2

2

1

2

1

2

1 ;N

N

i

i

N

N

v

vCircuit Representations

REFLECTING IMPEDANCES

dots)at signs (both 2

1

2

1

N

N

V

V

r)transforme leaving I(Current 21

2

2

1

N

N

I

I

Law) s(Ohm' 22 IZV L

2

11

1

21 N

NIZ

N

NV L 1

2

2

11 IZ

N

NV L

LZN

NZ

I

V2

2

11

1

1

sideprimary the into

reflected , impedance, LZZ 1

For future reference

2*22

*

1

22

2

12

*111 SIV

N

NI

N

NVIVS

ratio turns 1

2

N

Nn

21

21

21

21

SSn

ZZ

nIIn

VV

L

Phasor equations for ideal transformer

LEARNING EXAMPLEDetermine all indicated voltages and currents

25.04/1 n

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

21n

ZZ L

LZ

16321 jZ

5.1333.25.1342.51

0120

1250

01201 jI

1 1 1 (32 16) 2.33 13.5

83.36 13.07

V Z I j

12 14 (current into dot)

II I

n

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 nVVnVV

VV

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 Inn

VV SOC

2

Thevenin impedance will be the thesecondary mpedance reflected intothe primary circuit

22

n

ZZTH 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

02444

4

j

jVd

)4||4(2 jZTH

14.42 33.69 ( )OCV V

)(24 jZTH44

1688

44

162

j

jj

j

jZTH

2

48

1

1

1

62 j

j

j

j

jZTH

j

1

9024

This equivalent circuit is now transferred tothe secondary

LEARNING EXAMPLE (continued…)

Thevenin equivalent of primary side

2n

69.3384.28520

2

jVo

04.1462.20

69.3384.282

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

nIIn

VV

Phasor equations for ideal transformer

022

0101

211 I

VVV :1 Node @

022 2212

I

j

VVV :2 @Node

4 equations in 4 unknowns!

2n21

12

221

121

2

2

02)1(

01022

II

VV

IVjV

IVV

051I

05.22I

05)2)(1( 11 VjV

43.6324.2

05

21

051 jV 43.635

43.63522V