Magnetically Coupled Networks

1

2

Magnetically Coupled Networks

• A new four-terminal element, the transformer, is introduced in this chapter

• A transformer is composed of two closely spaced inductors, that is, two or more magnetically coupled coils– primary side is connected to the source– secondary side is connected to the load

3

dt

diMv 1

2

dt

diMv 2

1

4

Dot Convention• dot convention: dots are placed beside each

coil (inductor) so that if the currents are entering (or leaving) both dotted terminals, then the fluxes add

• right hand rule says that curling the fingers (of the right hand) around the coil in the direction of the current gives the direction of the magnetic flux based on the direction of the thumb

• We need dots on the schematic to know how the coils are physically oriented out one another

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Mutually Coupled CoilsThe following equations define the coupling between the two inductors assuming that each respective current enters the dot side which is also the positive voltage side where L1 and L2 are the self-inductances of the coils (inductors), and M is the mutual inductance between the two coils

dt

idL

dt

idMtv

dt

idM

dt

idLtv

22

12

2111

)(

)(

i1(t)

+

–

v1(t) L1

i2(t)

+

–

v2(t)L2

M

7

EXAMPLEWrite a set of mesh-current equations that describe the circuit shown in terms of the currents i1, i2, and i3

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(In the following set of mesh-current equations, voltage drops appear as positive quantities on the right-hand side of each equation.)Summing the voltages around the first mesh yields

The second mesh equation is

The third mesh equation is

dt

diii

dt

diivg

23121 5.498

1232312 865.440 iiiiii

dt

d

dt

di

3232

13 2065.490 iiidt

diii

dt

d

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Mutually Coupled Coils (AC)

• The frequency domain model of the coupled circuit is essentially identical to that of the time domain

2212

2111

IIV

IIV

LjMj

MjLj

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Source Input ImpedanceLinear Transformer

The source sees an input impedance, Zi, that is the sum of the primary impedance, and a reflected impedance, ZR, due to the secondary (load) side

LPRPP

Si f ZZZZ

IV

Z

L1 L2

M

ZL+–

VS

Z

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Transformer

source load

THE LINEAR TRANSFORMER

ZS

ZLI1

I2

R1 R2

jωL1jωL2

Vs

a

b

c

d

jωM

source Load

12

2221

2111

0 IZLjRMIj

MIjILjRZV

L

ss

L

s

ZLjRZ

LjRZZ

2222

1111

122

222211

2

222211

221

IZ

MjV

MZZ

MjI

VMZZ

ZI

s

s

13

22

22

1122

222211

int1 Z

MZ

Z

MZZZ

I

Vs

L

sab

ZLjR

MLjR

ZZ

MZZ

22

22

11

22

22

11

14

REFLECTED IMPEDANCE

LLL jXRZ

LL

LL

LL

LLr

XLjRRZ

M

XLRR

XLjRRM

XLjRR

MZ

222

22

22

22

22

2222

22

22

LL XLjRRZ 2222

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The parameters of a certain linear transformer are R1 = 200 Ω,R2 = 100 Ω, L1 = 9 H, L2 = 4 H, and k = 0.5. The transformercouples an impedance consisting of an 800 Ω resistor in series with a 1 µF capacitor to a sinusoidal voltage source. The 300 V (rms) source has an internal impedance of 500 + j100 Ω and a frequency of 400 rad/s.

a) Construct a frequency-domain equivalent circuit of the system.b) Calculate the self-impedance of the primary circuit.c) Calculate the self-impedance of the secondary circuit.d) Calculate the impedance reflected into the primary winding.e) Calculate the scaling factor for the reflected impedance.f) Calculate the impedance seen looking into the primary terminals of the transformer.g) Calculate the rms value of the primary and secondary current.h) Calculate the rms value of the voltage at the terminals of the load and source.i) Calculate the average power delivered to the 800 Ω resistor.j) What percentage of the average power delivered to the transformer is delivered to the load?

EXAMPLE

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S 0 L U T I 0 N

a) frequency-domain equivalent circuit of the system

;3495.0

;16004400

;36009400

2

1

HM

jjLj

jjLj

.2500400

101

;120034006

jjCj

jjMj

17

3700700360020010050011 jjjZ

c). The self-impedance of the secondary circuit.

900900250080060010022 jjjZ

d). The impedance reflected into the primary winding.

e). The scaling factor by which Z22* is reflected is 8/9

800800900900

9

8 900900

900900

12002

jjjj

Z r

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f). The impedance seen looking into the primary terminals of the transformer is the impedance of primary winding plus the reflected impedance, thus

440010008008003600200 jjjZab

g). Calculate I1 and I2

rmsmAj

jZZ

VI

abs

s

0

0

1

57.7125.636020

45001500

0300

rmsmAIj

jI 0

12 43.6363.59900900

1200

19

WIP

isloadthetodeliveredpoweraverageThei

rmsVIjIZV

rmsVIjV

arertransformeofterminalstheatvoltagesTheh

ab

84.2800

).

63.538.28544001000

82.852.1562500800

).

2

2

0111

022

20

WIPab 00.410002

1

j). The average power delivered to the transformer is

Therefore

%11.7110000.4

84.2 x

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Energy Analysis

• An energy analysis of the mutually coupled inductors provides an expression for the instantaneous stored energy

• The sign is positive (+) if currents are both entering (or leaving) the dots; sign is negative (-) if currents are otherwise

)()()()()( 212

22212

1121 titiMtiLtiLtw

22

212222

12112

1 iMiiLiLtw

0212222

12112

1 iMiiLiL

022 2121

2

22

11

MLLiii

Li

L

MLL 21

10 21 kLLkM

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Quantifying the Coupling• The mutual inductance, M, is in the range

• The coefficient of coupling (k) between two inductors is defined as

– for k > 0.5, inductors are said to be tightly coupled– for k 0.5, coils are considered to be loosely coupled

210 LLM

1021

LL

Mk

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The self-inductances of the coils shown are L1 = 5 mH and L2 = 33.8 mH. If the coefficient of coupling is 0.96, calculate the energy stored in the system in millijoules when (a) i1 = 10 A, i2 = 5 A; (b) i1 = -10 A, i2 = -5 A; (c) i1 = -10 A, i2 = 5 A; and (d) i1 = 10 A, i2 = -5 A.

DRILL EXERCISE