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10. Magnetically coupled networks

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1. Transformer

① Used for changing AC voltage levels.

② Transmission line : high voltage levels are used to decrease power loss due to

resistance of copper wires. The smaller magnitude of current, the less power loss,

when transmitting the same power.

③ Used for Impedance matching. Transformers are used to change magnitudes of

impedances to achieve maximum power transfer condition.

2. Inductors or transformers are difficult to integrate in an IC. (occupies large areas)

Transformer, Inductor

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Two important laws on magnetic field

Current generates magnetic field (Biot-Savart Law)

inducedV

Time-varying magnetic field generates induced electric field that opposes the variation. (Faraday’s law)

Current

Current

B-field

Top view

Electric field

B-field

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Current B i

l

NB

length

lengthl

Ni

il

NSBSadB

lengthS

Magnetic flux :

Magnetic flux

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inducedV

Current

,B

dt

dNVinduced

dt

di

l

SN

dt

dNV

lengthinduced

2

dt

diLVinduced

lengthl

SNL

2

2NL

Self inductance

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Inductor circuit

Magnetic fieldTotal magnetic flux linked by N-turn coil

Ampere’s Law(linear model)

Faraday’sInduction Law

Ideal Inductor

Assumes constant L and linear models!

SNi

dt

dN

dt

idL

dt

idNl

S

dt

dN

2

(S : cross-section area of a coil, μ : permeability)

dt

idL

The current flowing through a circuit induces magnetic field (Ampere’s law). A sudden change of a magnetic field induces electric field that opposes the change of a magnetic field (Faraday’s law), which appear as voltage drops across an inductor terminals.

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Mutual Inductance

dt

diL

dt

diL

N

Ni

N

L

dt

dN

dt

dN 1

211

11

21

1

1222

(1) When the secondary circuit is open

The current flowing through the primary circuit generates magnetic flux, which influences the secondary circuit. Due to the magnetic flux, a repulsive voltage is induced on the secondary circuit.

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Primary circuitSecondary circuit

Nomenclatures

Primary coilSecondary coil

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9Secondary voltage and current with different coil winding directions

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Two-coil system (both currents contribute to flux)

Self-inductance

Mutual-inductance

(From reciprocity)

(2) Current flowing in secondary circuit

dt

diL

dt

diL

dt

d 212

11

11

MLL 2112

dt

diL

dt

diL

dt

d 121

22

22

Self-inductance

Mutual-inductance

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The ‘DOT’ Convention

Dots mark reference polarity for voltages induced by each flux

dt

diM

dt

diL

dt

d 211

11

dt

diM

dt

diL

dt

d 122

22

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Example 10.2

Mesh 1 Voltage terms

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Mesh 2 Voltage Terms

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Example 10.4 0 VOLTAGE THE FIND V

1V

2V

2I

1123024 VI :KVL

022 222 IIjV- :KVL

)(62

)(24

circuit inductance Mutual

212

211

IjIjV

IjIjV

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

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Example E10.3 WRITE 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

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)(13 jZS )(11 jZL

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

DETERMINE IMPEDANCE SEEN BY THE SOURCE

?1

I

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

Example 10.6

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10.2 Energy analysis

)()()(

)()()( 121

1111 tidt

tdiM

dt

tdiLtittp

)(

)()()()()( 2

122222 ti

dt

tdiM

dt

tdiLtittp

0)()(2)()(2

1

)()(2)()(2

1

)()()(

)()()(

)()()()()()(

121112221

211

0

21222

211

0

212

2121

1

0

2211

0

21

1

1

11

tItIMtILtIL

dttitiMtiLtiLdt

d

dttidt

tdiM

dt

tdiLti

dt

tdiM

dt

tdiL

dttittitdttptpW

t

t

tt

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21

22

1

2

2

2

21

11

121112221

211

0

0)()(2)()(

LLM

IL

MLI

L

MIL

tItIMtILtIL

)10(21

21

k

LL

Mk

LLkM

; Coefficient of coupling

Coupling coefficient

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19Compute 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(

Example 10.7

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10.3 The ideal transformer

Insures that ‘no magnetic fluxgoes astray’

11 N 22 N

First ideal transformerequation

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

1

2

2

1

N

N

i

i Second ideal transformer

equations

Circuit Representations

1

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

2

1

2

1

22

11

)()(

)()(

N

N

v

v

tdt

dNtv

tdt

dNtv

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Reflecting Impedances

dots)at signs (both 2

1

2

1

N

N

V

V

er) transformleaving I(Current )( 2

1

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

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Non-ideal transformer

2

12122

2

12122

2

2

11

1

12

2

111

12

2122

122222

2111

21

)1()()(

)(

)(

)(,0

LjZ

ZLjLLk

LjZ

ZLjLLM

LjZ

MLj

I

VZ

ILjZ

MILjV

ILjZ

MjIMIjILjZ

MIjILjVVIZ

MIjILjV

LLkM

L

L

L

L

Lin

L

LL

L

LLin

L

LL

in

ZN

NZ

L

LZ

LjZ

ZLjZ

Lj

I

VZ

k

2

2

1

2

1

2

2

1

1

1

)2(

1)1(

To build ideal transformers, following two conditions are needed.

(1) k=1;(2) ZL<<jωL;

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Example 10.8 Determine 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!

LZ

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Thevenin’s equivalents 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’

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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

Thevenin’s equivalents from primary

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Draw the two equivalent circuits

Equivalent circuit reflectinginto primary

Equivalent circuit reflecting into secondary

2n

Example 10.9

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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

Example E10.8

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oVFind

Transfer tosecondary

2n

024

88 j

04

2

OV

0202)88(

2

jVO

66.3881.12

040OV

Example E10.9

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Safety considerations

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