MAGNETICALLY COUPLED NETWORKS

MAGNETICALLY COUPLED NETWORKS

LEARNING GOALS

Mutual Inductance Behavior of inductors sharing a common magnetic field

Energy Analysis Used to establish relationship between mutual reluctance and self-inductance

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

Safety Considerations Important issues for the safe operation of circuits with transformers

BASIC CONCEPTS – A REVIEW

Magnetic field Total magnetic flux linked by N-turn coil

Ampere’s Law(linear model)

Faraday’sInduction Law

Ideal Inductor

Assumes constant L and linear models!

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( )

What happens if the flux created by thecurrent links to another coil?

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

j

i

i

ij

i

n

jiji

circuit in

current aby caused circuit linkingFlux

circuit linkingflux Total

1

Assume n circuits interacting

jijij iLmodelsinductor linear For

ji

LL

iL

jiij

ii

and

circuits between inductance Mutual

circuit of "inductance self"

Special case n=2

2221212

2121111

iLiL

iLiL

2112 LL :ModelLinear

A GENERALIZATION

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 Voltage terms

Mesh 2

LEARNING EXAMPLE - CONTINUED

Voltage Terms

1i

1v

dt

diL

dt

diMv

dt

diM

dt

diLv

22

12

2111

)()()(

)()()(

22

12

2111

tdt

diLt

dt

diMtv

tdt

diMt

dt

diLtv

Equivalent to a negative mutualinductance

2i

2vdt

diM

dt

diLv 21

11

dt

diL

dt

diMv 2

21

2

More on the dot convention

LEARNING EXTENSION

)()()( 211 t

dt

diMt

dt

diLtv

)()()( 22

12 t

dt

diLt

dt

diMtv

)()()( 211 t

dt

diMt

dt

diLtv

)()()( 22

12 t

dt

diLt

dt

diMtv

Convert tobasic case

)(),( 21 tvtvfor equations the Write

PHASORS AND MUTUAL INDUCTANCE

)()()(

)()()(

22

12

2111

tdt

diLt

dt

diMtv

tdt

diMt

dt

diLtv

2212

2111

ILjMIjV

MIjILjV

Phasor model for mutually

coupled linear inductorsAssuming complex exponential sources

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

CASE I

1V 2V ILjMIjV

MIjILjV

22

11

Currents into dots

CASE 2

1V 2V

Currents into dots

I I

I I 21 VVV

21 VVV

ILjMIjV

MIjLjV

22

11

ILjMLLjV eq )2( 21

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

1I 2I 21 III

221

211

ILjMIjV

MIjILjV

I

MLL

MLLV

221

221

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

VS

816

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 EXTENSION021 ,, VII FIND

1V

2V

1. Define variables for coupled inductors

2. Loop equations

SV

04 11 IVVS

0)42( 22 IjV

212

211

8

4

IjjIV

jIIjV

3. Coupled inductors equations

0)42(

)44(

21

21

IjjI

VjIIj S

4. Replace and rearrange

20 4IjV

)44/(

/

j

j

SjVIjj 2))21)(1(81(

j

jVI S

2472

j724

024

j

j

26.1625

024

)(26.1696.02 AI

Voltages in VoltsImpedances in OhmsCurrents in ____

2121 )42(/0)42( IjjIjIjjI

26.1696.043.6347.49011I

)(17.13729.41 AI

26.1696.049014 20 IjV

)(26.10684.30 VV

LEARNING EXTENSIONWRITE THE KVL EQUATIONS

1. Define variables for coupled inductors

1I 2I

aV bV

2. Loop equations in terms of inductor voltages

0)( 111212 IRVIIRVa

0)( 122123 IIRVIRVb

3. Equations for coupled inductors

)( 211 IMjILjVa )( 221 ILjMIjVb

4. Replace into loop equations and rearrange

1221121 VIMjRILjRR

1223212 VILjRRIMjR

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

jj

1

133

j

j

1

1

)(5.25.32

133 j

jjZi

)(54.3530.4 iZ

WARNING: This is NOT a phasor

LEARNING EXTENSIONDETERMINE IMPEDANCE SEEN BY THE SOURCE

)(12 jZS

)1||2(2 jjZL

)(222

2

jj

ZL

4. Replace and do the algebra

2

2

11

)()(

LjZ

MjLjZ

I

VZ

LS

Si

jj

jjjZ i 2)22(

)1(2)12(

2

)21(2

1)2(

jj

)21(

)21(

j

j

)(8.01.2)21(2

212 2

jj

jZ i

)(85.2025.2 iZ

022

11

IZV

VIZV

L

SS

2212

2111

ILjMIjV

MIjILjV

1I

1V

2V

2I

1. Variables for coupled inductors

One can choose directionsfor currents. If I2 is reversed one getsthe same equations thanin previous example.Solution for I1 must be thesame and expression forimpedance must be the same

2. Loop equations 3. Equations for coupled inductors

SV

ENERGY ANALYSISWe determine the total energy stored in a coupled network

This development is different from the one in the book. But the finalresults is obviously the same

)()()(

)()()(

22

12

2111

tdt

diLt

dt

diMtv

tdt

diMt

dt

diLtv

INDUCTORS COUPLED FOR EQUATIONS

)()()()()( 2211 titvtitvtpT NETWORK TO SUPPLIED POWERTOTAL

)(/

)(/

2

1

ti

ti

)()()()(

)()()()()(

2222

1

21

111

tdt

ditiLtit

dt

diM

tdt

ditMit

dt

ditiLtpT

)(21 tdt

idiM )(

2

1 22 tdt

di)(

2

1 21 tdt

di

)(

2

1)()()(

2

1)( 2

2221211 tiLtitMitiL

dt

dtpT

t

)(2

1)()()(

2

1)( 2

2221211 tiLtitMitiLtw

2

12

2221

2

2

1 )()(2

1)(

2

1)(

ti

L

MtiLti

L

MLtw

)(2

1)(

2

1 21

2

221

2

2

tiL

Mti

L

M

0)(2

2

1 L

MLtw 21LLM

21LL

Mk Coefficient of

coupling

LEARNING EXAMPLECompute the energy stored in the mutually coupled inductors

1k

mHLmHL 61.10,653.2 21

mst 5

)(2

1)()()(

2

1)( 2

2221211 tiLtitMitiLtw

Assume steady state operation

We can use frequency domain techniques

2121 ,, LLkMkLL

)(),(, 21 titiM COMPUTE MUST

mHM 31.5

110653.2377 31L

2,42 ML

Circuit in frequency domain

1I 2I

Merge the writing of the loop and coupledinductor equations in one step

0)42(4

024)21(2

212

211

IjIjI

IjIjI

)(69.3333.3),(31.1141.9 21 AIAI GET TO SOLVE

))(69.33377cos(33.3)(

))(31.11377cos(41.9)(

2

1

Atti

Atti

radians! in is term The :WARNING t377 108)(885.1377005.0 radtst

)(61.2)005.0(),(10.1)005.0( 21 AiAi

)()61.2(1061.1015.0

)61.2()10.1(1031.5

)10.1(10653.25.0)005.0(

23

3

23

J

w

mJw 5.22)005.0(

LEARNING EXTENSION mst 10 ATSTOREDENERGY DETERMINE

Hzf 60

)(2

1)()()(

2

1)( 2

2221211 tiLtitMitiLtw

3012)12(2 211 IjIjI0)22()21( 221 IjIjIj

02

3012)22(

21

21

IjI

jIIj

3012)5.022( 1Ij

12 5.0 jII

66.3820.3

3012

25.2

30121 jI )(66.875.3 A

66.98875.1

66.875.3905.05.0 12 jII

)(9.37860 1 sHzf

))(66.989.378cos(875.1)(

))(66.89.378cos(75.3)(

2

1

Atti

Atti

1.217)(789.3(sec)010.0sec)/(9.378 radrad

)(91.0)010.0(

)(3.3)010.0(

2

1

Ai

Ai

)()91.0(*00528.0*5.0

)91.0)(3.3(*00264.0

)3.3(*00528.0*5.0)010.0(

2

2

J

w

211 )(00528.02 LHLL )(00264.0 HM

Go back to time domain

)91.0)91.0)(3.3(3.3(*00264.0)010.0( 22 w

mJJw 30030.0)010.0(

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

Circuit Representations1

2

2

1

2

1

2

1 ;N

N

i

i

N

N

v

v

Since the equationsare algebraic, theyare unchanged forPhasors. Just becareful with signs

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

012021

1111 ZZ

ZIZV

1205.1342.51

16320120

5.1333.2)1632(

21

1

11

j

ZZ

Z

jIZ SAMECOMPLEXITY

07.1336.835.1333.257.2678.351V

dot) into(current 11

2 4In

II

dot) to opposite is ( 112 25.0 VnVV

CAREFUL WITH POLARITIES ANDCURRENT DIRECTIONS!

LEARNING EXTENSION 1Icurrent the Find

2n

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

21n

ZZ L

1Z

5.014

241 j

jZ

5.23 jZ i

5.23

0121 jI )(81.3907.3

81.3991.3

012A

Voltage in VoltsImpedance in Ohms...Current in Amps

LEARNING EXTENSION oV Find

Strategy: Find current in secondary and then use Ohm’s Law

n

II 1

2

)(81.3907.32

22

10

12 V

IV

II

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

Equivalent circuit reflectinginto primary

Equivalent circuit reflecting into secondary

2n

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

)(69.3347.14 VVOC

)(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

Circuit with primary transferred to secondary

69.3384.28520

2

jVo

04.1462.20

69.3384.282

Transfer tosecondary

LEARNING EXTENSION 1I Find

Equivalent circuit reflectinginto primary

2n

5.0)(22 j

036

1I

5.12

060361 jI

86.365.2

0361I

2

012 Notice the position of thedot marks

LEARNING EXTENSION oVFind

Transfer tosecondary

2n

024

88 j

04

2

OV

0202)88(

2

jVO

66.3881.12

040OV

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 j

V 43.635

43.63522V

SAFETY CONSIDERATIONS: AN EXAMPLE Houses fed from different distributiontransformers

Braker X-Y opens, house Bis powered down

Good neighbor runs an extensionand powers house B

When technician resets thebraker he finds 7200V betweenpoints X-Z

when he did not expect to find any

LEARNING BY APPLICATIONWhy high voltage transmission lines?

CASE STUDY: Transmit 24MW over 100miles with 95% efficiencyA. AT 240VB. AT 240kV

2

5

8

1060919160

108

rA

m.Km.l

A

l

section cross

conductor of length

/m)copper (e.g., material ofy resistivit

R ,resistanceConductor :Given

MWMWPloss 2.105.024 losses, Maximum :Required

AV

WI l

56

10240

1024

current a needs one240V At A.

2RIPloss :losses Line :Known

102

586 10

10609.11082102.1

rW

mr 624.8

AV

WI l

23

6

1010240

1024

current a needs one240kV At B.

42

586 10

10609.11082102.1

rW

cmr 8624.0

LEARNING EXAMPLERating a distribution transformer

A200 householdper current Maximum

10 r transformeper Households

nV

V

1

2

5.57

1

13800

240nDetermining ratio

Determining power rating

2

1

2000

I

In

A current secondary Max

AI 78.346.572000

1

kVAP 48078.34800,13

)(2000)(240 AVP :Also

LEARNING EXAMPLEBattery charger using mutual inductance

Assume currents in phase

100RMS

mA0.5RMS

A

Smaller inductor

BATTERY CHARGER WITH HIGH FREQUENCY SWITCH

APPLICATION EXAMPLEINDUCED ELECTRIC NOISE

“NOISE” 0.1

10AC DC

L L nH

CASE 1: AC MOTOR (f=60Hz)1.88NOISEV V

CASE 2: FM RADIO TRANSMITTER

3.14NOISE

V V !!!

LEARNING EXAMPLELINEAR VARIABLE DIFFERENTIAL TRANSFORMER (LVDT)

10

2IN RMS

V V

f kHz

NO LOAD!

12 13

2

| | 0.8

| | | |

1

0.8

O IN

S

P

k k

V V

L

L

LVDT - CONTINUED

Complete AnalysisAssuming zero- phase input

21.25 1.5625S

P

L

L

FOR EXAMPLE

10

10

2000

P RMS

IN RMS

I mA

V V

f Hz

Design equations forinductances

LEARNING BY DESIGNUse a 120V - 12V transformer to build a 108V autotransformer

Conventional transformer Auto transformer connections

yzxyxz VVV xzxyxz VVV

Circuit representations

Use the subtractive connectionon the 120V - 12V transformer

Transformers

DESIGN EXAMPLE DESIGN OF AN “ADAPTOR” OR “WALL TRANSFORMER”

2

2

| | 12

0.6

DCP W

V

Design constrains and requirements

Transformer turn ratio

Notice safety margin

25 : 2

Specify 100mA rating for extramargin!