14 indutancia

7/30/2019 14 Indutancia

1/33

Inductances and

InductorsDragica Vasileska

ProfessorArizona State University


2/33

Flux Linkage

Consider two magnetically coupled circuits

C1I1

S1 S2 C2

I2


3/33

Flux Linkage (Contd)

The magnetic flux produced I1 linking thesurface S2 is given by

If the circuit C2 comprisesN2 turns and thecircuit C1 comprisesN1 turns, then the totalflux linkage is given by

2

2112S

sdB

2

2121122112

S

sdBNNNN


4/33

Mutual Inductance

The mutual inductance between two circuits

is the magnetic flux linkage to one circuitper unit current in the other circuit:

1

1221

1

1212

I

NN

I

L


5/33

Neumann Formula for Mutual

Inductance

2

2

21

1

21

21

1

21

1

1221

1

1212

C

S

ldAI

NN

sdB

I

NN

I

NN

I

L


6/33

Neumann Formula for Mutual

Inductance (Contd)

1

12

1101

4C

R

ldIA

1 2

2

12

21210

21

1

2112

4C C

C

R

ldldNN

ldAI

NNL


7/33


8/33

Self Inductance

Self inductance is a special case of mutual

inductance.

The self inductance of a circuit is the ratio of the

self magnetic flux linkage to the currentproducing it:

1

11

2

1

1

1111

IN

IL


9/33

Self Inductance (Contd)

For an isolated circuit, we call the selfinductance, inductance, and evaluate itusing

I

N

I

L

2


10/33

Self-Inductance

Formula by Definition

Applies to linear magnetic materials only

Units:

flux linkage

current through each turn

mN

LI

2[Henry] [H] [Wb/A] [T m /A]L


11/33

Inductance of Coaxial Cable

Magnetic Flux

Inductance

(as commonly used in transmission line theory)

0

( ) ( )2

ln( / )2 2

m

S S

b d

a

IB dS d dz

I Idd dz b a

ln( / ) [H]

2

or ln( / ) [H/m]2

md

L b a

IL

b ad


12/33

Inductance of Toroid

Magnetic FluxDensity

Magnetic Flux

If core small

vs. toroid

2[T] [Wb/m ]2

NIB

m

S

B dS

2

0

0

(if )2

where S cross section area of the toroid core

mB S

NISS


13/33


14/33

Alternative Approaches

Self-inductance in terms of

Energy

Vector magnetic potential (A)

Estimate by Curvilinear Square Field Map method

2

2

21

2

H

H

WW LI L

I


15/33

Energy Stored in Magnetic Field

The magnetic energy stored in a regionpermeated by a magnetic field is given by

dvHdvHBW

VV

m 2

2

1

2

1


16/33

Energy Stored in an Inductor

The magnetic energy stored in an inductoris given by

2

2

1LIW

m


17/33

Inductance of a

Long Straight Solenoid

Energy Approach

Inductance

2

. .

2 2 2 2

2 2 0. ( )

2 2

2

1

2 2

where for this solenoid

2 2

where for a circular core2

H

vol vol

d

H

vol S core

H

W B Hdv H dv

NIHd

N I N IW dv dS dz

d d

N I SW S ad

2

2

2H

W N SL

I d


18/33

Internal Inductance

of a Long Straight Wire

Significance: an especially important issuefor HF circuits since

Energy approach (for wire of radius a)

L L LZ X L Z

2

2

. .

22

3

2 4 0 0 0

2 24

2 4

1( )

2 2 2

8

( / 4)(2 )( )

8 16

H

vol vol

a l

IW B Hdv d d dz

a

Id dz

a

I I la l

a


19/33

Internal Inductance

of a Long Straight Wire

Expressing Inductance in terms of energy

Note: this result for a straight piece of wire

implies an important rule of thumb for HFdiscrete component circuit design:

keep all lead lengths as short as possible

2

2 2

2( )2 16

8

or8

H

I l

W lL

I IL

l


20/33

Example of Calculating

Self-Inductance

Exercise 1

Find: the self-inductance of

a) a 3.5 m length of coax cable with a = 0.8 mm

and b = 4 mm, filled with a material for which

r = 50.

0

7

ln( / ) ln( / )

2 2(50)(4 10 H/m)(3.5m) 4

ln( )2 0.8

56.3 H

rdd

L b a b a

L


21/33


Self-Inductance

Exercise 1 (continued)

Find: the self-inductance of

b) a solenoid having a length of 50 cm and 500 turns

about a cylindrical core of 2.0 cm radius in which r =50 for 0 < < 0.5 cm and r = 1 for 0.5 < < 2.0 cm

2 2 22 2

0

2 6 2

2 2 3 2

6 3

( ) (50 )

where (.005 m) 78.5 10 m

and [(.020 m) (.005 m) ] 1.18 10 m

(4 10[(50)(78.5 10 ) 1.18 10 ]

i i o o

i i o o i o

i

o

N S N S NN S NL S S S S

d d d d d

S

S

L

7 2)(500)3.2 mH

0.50


22/33

Inductors in Series and in

Parallel Inductors, like resistors and capacitors, can be

placed in series

Increasing levels of inductance can be obtained by

placing inductors in series


23/33

Inductors in Series and in Parallel

Inductors, like resistors and capacitors, can beplaced in parallel.

Decreasing levels of inductance can be obtained by

placing inductors in parallel.


24/33

Mutual Inductance

Significant when current in one conductorproduces a flux that links through the pathof a 2nd separate one and vice versa

Defined in terms of magnetic flux (m)

2 12

121

12 1 2

2

mutual inductance between circuits 1 and 2

where the flux produced by I that links the path of I

and N the # of turns in circuit 2

NM

I


25/33

Mutual Inductance

Expressed in terms of energy

Thus, mutual inductances betweenconductors are reciprocal

12 1 2 0 1 2

1 2 1 2. .

12 21

1 1

and [H/m]

vol vol

M B H dv H H dv

I I I I

M M


26/33


Mutual Inductance

Given: 2 coaxial solenoids, eachl= 50 cm

long

1st: dia. D1= 2 cm, N1=1500 turns, core r=75

2nd: dia. D2=3 cm, N2=1200 turns, outside 1st

Find: a) L1=? for the inner solenoid2 22

0 1 11 1 1

1

7 2 2

1

4

(75)(4 10 H/m)(1500) (.02m)

4(.50m)

.133 H = 133 mH

rN DN S

L l l

L


27/33


Mutual Inductance

Continued

Find: b) L2 = ? for the outer solenoid

Note: this solenoid has inner core and outer air

filled regions as in Exercise 1 part c), soit may be treated the same way!

2 0.087 H 87 mHL


28/33


Mutual Inductance

Continued

Find: M = ? between the two solenoids

1

2 12 2 1 112 12

1

7 2

1 2

using since core 1 is smaller of the two

(75)(4 10 )(1200)(1500) (.02)

4(.50)107 mH

( geometric mean of the self-inductance

of each

S

N N N S

M M MI l

M

M

L L

individual solenoid)


29/33

Mutual Inductance Between

Circular Loops A circular loop of conducting wire

of radius acarries current I. Findthe magnetic field on the axis ofthe loop a distance h from theplane of the loop by directintegration of the Biot-Savart Law.

If a small circular circuit of radiusis placed at this position (so thatthe magnetic field may beconsidered uniform over the area ofthe small loop) such that the planes

of the two circuits are parallel, findthe mutual inductance betweenthem.


30/33

Solution

The element of magnetic field at distance halong theaxis, due to a current element is:

The components of the various along the axis all add,while those normal to the axis sum to zero. Themagnitude of the component of along the axis is:


31/33

So the total field along the axis is:

The magnetic flux through the loop of radius(normal to ) is:

Since the mutual inductance M is defined by :


32/33

Summary

Inductance results from magnetic flux(m) generated by electric current in aconductor

Self-inductance (L) occurs if it links with itself Mutual inductance (M) occurs if it links with

another separate conductor

The amount of inductance depends on How much magnetic flux links

How many loops the flux passes through

The amount of current that generated the flux


33/33

Summary

Inductance formulas may be derived from Direct application of the definition

Energy approach

Vector Potential Method

The self-inductance of some common structureswith sufficient symmetry have an analytical result

Coaxial cable Long straight solenoid

Toroid

Internal Inductance of a long straight wire

14 indutancia

Documents