Electrical Circuits II (ECE233b)

The University of Western Ontario Faculty of Engineering Science

Anestis Dounavis

Magnetically Coupled Networks

Mutual Inductance

+V-

i

N

Ampere’s law predicts that the flow of electric current will create a magnetic filed.

Faraday’s law predicts that time varying magnetic field will create a voltage potential.

dtdv t

where vt is the voltage per turn induced by a time-varying magnetic flux , through a coil.

A coil having N turns induces a total voltage of

dtdλ

dtdNNvv t

where is called the flux linkage and =N (weber-turn).

Mutual Inductance

+V-

i

N dtdλ

dtdNNvv t

The flux linkage induced by the current i has the following relationship in a constant permeable medium Liλ

where L is a constant, referred to as the inductance (henry).

Nλ ;

The voltage induced can know be written as

dtdiL

dtdλv

L

i+V-

or if time variation is sinusoidal

LIjV (phasor representation)

Mutual Inductancei1

+v1

-

i2+v2

-2211

12 21

If the magnetic field that induces the current is produced by some external source (i.e. adjacent circuit) then this produces a mutual inductance

There are no electrical connections between adjacent circuits A portion of the magnetic flux is shared between the coils

The total fluxes are 2121111 iLiLλ 2221212 iLiLλ

dtdiM

dtdiL

dtdλv 21

11

1

For linear circuits: MLL 2112 111 LL 222 LL & substituting

dtdiL

dtdiM

dtdλv 2

212

2

Voltages of coupled network becomei1

+v1

-

i2+v2

-

Circuitdiagram

M

L1 L2

i1 i2+v1

-

+v2

-

Polarity (dot convention)Dot convention indicates if 11 and 22 are added or subtracted

Circuit equations for voltages and currents adhere to passive element sign convention

ExampleR

+v-

i

v = iR R+v-

i

v = -iR

If reference current enters the dot, the induced voltage is positive at the dot in the other coil.

If reference current leaves the dot, the induced voltage is negative at the dot in the other coil.

L1 L2

M

Polarity (dot convention) If reference current enters the dot, the induced voltage is positive at the dot in the other coil.

If reference current leaves the dot, the induced voltage is negative at the dot in the other coil.

+v1

-

+v2

-

i1 i2+v1

-

i1 L1

-+dt

diM 2

dtdiM 1

+v2

-

i2L2

-+

dtdiM

dtdiLv 21

11

dtdiL

dtdiMv 2

21

2

Case 2

L1 L2

+v1

-

i1 L1

+-dt

diM 2

dtdiM 1

+v2

-

i2L2

+-

dtdiM

dtdiLv 21

11

dtdiL

dtdiMv 2

21

2

Case 1

+v1

-

+v2

-

i1 i2

L1 L2

M

M

Example 1

-v1

+

+v2

-

i1 i2

L1 L2

MDetermine the equations for v1 and v2 for the circuit shown

Example 2

-v1

+

+v2

-

i1 i2

L1 L2

MDetermine the equations for v1 and v2 for the circuit shown

Example 3Two coupled coils can be interconnected in 4 possible ways (see Example 2 pp. 320). L1 L2

For the following interconnection determine the equivalent inductance

L1 L2

+ V -

Example 4

I1

-+24|0 0

4

j4 j8

j1 2

-j4+

Vo-

I2

Find the current I1 and I2 and the output voltage Vo, for the circuit shown

Example 5

Write the KVL equations in the standard form for the network shown

I1

+-

I2L1 L2

R1

R2R3

V1

Energy Analysis

i1 i2+v1

-

+v2

-

M

L1 L2

The energy stored in a circuit is defined as 1 1t

0

t

0

v(t)i(t)dtp(t)dtE

For magnetically coupled inductors, the total energy in a network can be expressed as (derivation pp 323-325)

21222

211 iMiiL

21iL

21E(t)

Or if one of the dot is at the bottom i1 i2+v1

-

+v2

-

M

L1 L221

222

211 iMiiL

21iL

21E(t)

Energy Analysis Therefore, the general form of the energy stored in the magnetically coupled inductors at any instant of time is

21222

211 iMiiL

21iL

21E(t)

The energy expression can be rewritten in matrix form as:

2

1

2

121 i

iLMML

ii21E(t)

The two coupled inductors represent a passive network, therefore the energy stored within this network must be nonnegative

This means that L1>0, L2>0 and the matrix

2

1

LMML

is nonnegative definite.

Energy Analysis

2

1

LMMLSince the matrix is nonnegative definite, then

the determinant of the matrix must also be nonnegative definite

0det 221

2

1

MLL

LMML

Hence for a passive network 21|| LLM

The coupling between the 2 inductors is defined as21LL

Mk

where k is defined as the coupling coefficient and ranges from10 kNo flux common to

the 2 coils (i.e. M=0)All flux common to 2 coils (no “leakage” flux)

In practice in transformers1k

Example 6

I1

+-12|30 0

2

j j2

j1 2

-j2

I2

The network shown operates at 60 Hz. Compute the energy stored in the mutually coupled inductors at time t=10ms.

Ideal TransformersAssume lossless magnetic core (no hysteresis or eddy currents and perfect coupling (k=1))

i(t)1 i(t)2+v(t)1

-

+v(t)2

-N1

N2

If the same flux goes through each winding thenFaraday’s law:

dtdNv 11

dtdNv 22

The ratio of these two equations yield

2

1

2

1

NN

vv

Ideal TransformersUsing Amperes Law (assuming ideal magnetic core with infinite permeability), it can be shown that

i1 i2+v1

-

+v2

-

N1:N2

L1 L2

0iNiN 2211 or1

2

2

1

NN-

ii

Equivalent Circuit Diagram

2

1

2

1

NN

vv

1

2

2

1

NN-

iiwhere

Usually the direction of i2 is often reversed: i1 i2

+v1

-

+v2

-

1:n

L1 L2 n1

NN

vv

2

1

2

1 nNN

ii

1

2

2

1 where

v1 source primary sourcev2 source secondary source

n turn ratio

Impedance TransformationConsider the following circuit: i1 i2

+v1

-

+v2

-

1:n

L1 L2ZLZ1

What’s the input impedance?

L2

2

21

11 Z

iv

iivZ

22

nnnv 11

2

n

nvv 2

1 21 nii 2

2L i

vZ Summary

2nL

1ZZ where n

NN

1

2

Note that the signs of these equations depend on the polarity w.r.t. the dot. For example:

i1 i2+v1

-

+v2

-

1:n

L1 L2ZLZ1 n

vv 21

21 nii

2nL

1ZZ

Voltages and currents are reversed (not the impedance)

Impedance Transformationi1 i2

+v1

-

+v2

-

1:nZ1 Z2

+-VS1

+- VS2

Consider the following circuit:

To simplify analysis the transformer and load can be replaced with equivalent circuitEquivalent Circuit Reflected to Secondary

i2i1+v1

-

+v2

-

1:nZ1

+-VS1

+vTH

-

i2 = 0, since far end terminals are open circuited

i1 = ni2 = 0vTH = v2 = nv1 = nVS1

i1+v1

-

+v2

-

1:nZ1

+-VS1

iN

i2v2 = nv1 = 0

1

S112N nZ

Vniii

Impedance Transformation

12

N

THTH Zn

ivZ

i1 i2+v1

-

+v2

-

1:nZ1 Z2

+-VS1

+- VS2

The original circuit

can be modeled with the following equivalent circuit

i2n2Z12 Z2

+-nVS1

+- VS2

Impedance TransformationEquivalent Circuit Reflected to Primary

i1 i2+v1

-

+v2

-

1:nZ1 Z2

+-VS1

+- VS2

In a similar manner we can replace the ideal transformer and its secondary circuit with an equivalent circuit as shown

i1Z1

+-VS1

+- (VS2)/n

22 /nZ

Example 7Compute the current I1 and the output voltage Vo for the circuit shown

I1

+-12|0 0

21:2

2

+Vo-

2

-j2-j2

+V1

-

+V2

-

Example 8For the circuit shown, find Vo

+-12|0 0

21:2

+Vo-

2

-j2

+ -

4|0 0

Example 9For the circuit shown, find I1

+-36|0 0

21:2

I1

2

-j2

+ -

12|0 0