Electromagnetics ENGR 367 Inductance. Introduction Question: What physical parameters determine how much inductance a conductor or component will have

Electromagnetics Electromagnetics ENGR 367ENGR 367

InductanceInductance

IntroductionIntroduction

►Question: What physical parameters Question: What physical parameters determine how much inductance a determine how much inductance a conductor or component will have conductor or component will have

in a in a circuit?circuit?

►Answer: It all depends on current and Answer: It all depends on current and flux flux linkages!linkages!

Flux LinkageFlux Linkage

►Definition:Definition: the magnetic flux generated by a the magnetic flux generated by a

current that current that passes through one or more passes through one or more conducting loops conducting loops of its own or another of its own or another separate circuit separate circuit

►Mathematical Expression:Mathematical Expression: If the total flux generated by N turns

and # of turns through which passes

then flux linkage (assuming none escapes)

m

m

m

N

N

Types of InductanceTypes of Inductance

►Self-Inductance (L):Self-Inductance (L):

whenever the flux linkage of a whenever the flux linkage of a conductor or circuit couples with itselfconductor or circuit couples with itself

►Mutual Inductance (M):Mutual Inductance (M):

if the flux linkage of a conductor if the flux linkage of a conductor or or circuit couples with another circuit couples with another separate oneseparate one

Self-InductanceSelf-Inductance

►Formula by DefinitionFormula by Definition

Applies to linear magnetic materials onlyApplies to linear magnetic materials only

Units:Units:

flux linkage

current through each turnmN

LI

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

Inductance of Coaxial CableInductance of Coaxial Cable

►Magnetic FluxMagnetic Flux

► InductanceInductance

(as commonly used in transmission line (as commonly used in transmission line theory)theory)

0

ˆ ˆ( ) ( )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

m dL b a

IL

b ad

Inductance of ToroidInductance of Toroid

►Magnetic Magnetic Flux Flux DensityDensity

►Magnetic Magnetic FluxFlux

► If core small If core small vs. vs.

toroidtoroid

2 [T] [Wb/m ]2

NIB

m

S

B dS

20

0

(if )2

where S cross section area of the toroid core

m B S

NISS

Inductance of ToroidInductance of Toroid


Result assumes that no flux escapes Result assumes that no flux escapes through through gaps in the windings (actual L gaps in the windings (actual L may be less)may be less)

In practice, empirical formulas are often In practice, empirical formulas are often used to used to adjust the basic formula for adjust the basic formula for factors such as factors such as winding (density) and winding (density) and pitch (angle) of the pitch (angle) of the wiring around the corewiring around the core

2

0

[H]2

mN N SL

I

Alternative ApproachesAlternative Approaches

► Self-inductance in terms ofSelf-inductance in terms of

EnergyEnergy

Vector magnetic potential (Vector magnetic potential (AA))

Estimate by Curvilinear Square Field Map Estimate by Curvilinear Square Field Map methodmethod

22

21

2H

H

WW LI L

I

Inductance of a Inductance of a Long Straight SolenoidLong Straight Solenoid

► Energy ApproachEnergy Approach


2

. .

2 2 2 2

2 2 0. ( )

2 22

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

NIH

d

N I N IW dv dS dz

d d

N I SW S a

d

2

2

2 HW N SL

I d

Internal Inductance Internal Inductance of a Long Straight Wireof a Long Straight Wire

►Significance: an especially important Significance: an especially important issue issue for HF circuits sincefor HF circuits since

►Energy approach (for wire of radius a)Energy approach (for wire of radius a)L L LZ X L Z

2

2

. .

2 232 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

Internal Inductance Internal Inductance of a Long Straight Wireof a Long Straight Wire

► Expressing Inductance in terms of energyExpressing Inductance in terms of energy

► Note: this result for a straight piece of wire Note: this result for a straight piece of wire implies an important rule of thumb for HF implies an important rule of thumb for HF discrete component circuit design: discrete component circuit design:

““keep all lead lengths as short as keep all lead lengths as short as possible”possible”

2

2 2

2( )2 168

or 8

H

I lW l

LI IL

l

Example of Calculating Example of Calculating Self-InductanceSelf-Inductance

► Exercise 1 (D9.12, Hayt & Buck, 7Exercise 1 (D9.12, Hayt & Buck, 7thth edition, p. edition, p. 298)298)

FindFind: the self-inductance of: the self-inductance of

a) a 3.5 m length of coax cable with a = 0.8 a) a 3.5 m length of coax cable with a = 0.8 mm mm and b = 4 mm, filled with a material for and b = 4 mm, filled with a material for whichwhich

rr = 50. = 50.0

7

ln( / ) ln( / )2 2

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

2 0.8 56.3 H

r ddL b a b a

L


► Exercise 1 (continued)Exercise 1 (continued)


b) b) a toroidal coil of 500 turns, wound on a a toroidal coil of 500 turns, wound on a fiberglass form having a 2.5 x 2.5 cm square fiberglass form having a 2.5 x 2.5 cm square cross section and an inner radius of 2.0 cmcross section and an inner radius of 2.0 cm

2 7 2 2

0

(1)(4 10 H/m)(500) (0.025m)

2 2 (0.020 m 0.0125 m)

0.96 mH

N SL

L


► Exercise 1 (continued)Exercise 1 (continued)


c) c) a solenoid having a length of 50 cm and 500 a solenoid having a length of 50 cm and 500 turns turns about a cylindrical core of 2.0 cm radius about a cylindrical core of 2.0 cm radius in in which which rr = = 50 for 0 < 50 for 0 < < 0.5 cm and < 0.5 cm and rr = 1 for 0.5 = 1 for 0.5 < < < 2.0 cm < 2.0 cm2 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 oi 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

Example of Estimating Example of Estimating Inductance: Structure with Inductance: Structure with

Irregular GeometryIrregular Geometry► Exercise 2: Exercise 2: Approximate the inductance per unit Approximate the inductance per unit

length of length of the irregular coax by the curvilinear the irregular coax by the curvilinear square methodsquare method

0

5.5

4(6)

(0.23)(400 nH/m)

(if filled with air or

non-magnetic material)

290 nH/m

S

P

NL

l N

L

l

Mutual InductanceMutual Inductance

►Significant when current in one Significant when current in one conductor produces a flux that links conductor produces a flux that links through the path of a 2through the path of a 2ndnd separate one separate one and vice versaand vice versa

►Defined in terms of magnetic flux (Defined in terms of magnetic flux (mm))2 1212

1

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

Mutual InductanceMutual Inductance

►Expressed in terms of energyExpressed in terms of energy

►Thus, mutual inductances between Thus, mutual inductances between conductors are reciprocalconductors are reciprocal

12 1 2 0 1 21 2 1 2. .

12 21

1 1

and [H/m]vol vol

M B H dv H H dvI I I I

M M

Example of Calculating Example of Calculating Mutual InductanceMutual Inductance

►Exercise 3 (D9.12, Hayt & Buck, 7/e, p. Exercise 3 (D9.12, Hayt & Buck, 7/e, p. 298)298) GivenGiven: 2 coaxial solenoids, each : 2 coaxial solenoids, each l l = 50 cm = 50 cm

long long

11stst: dia. D: dia. D11= 2 cm, N= 2 cm, N11=1500 turns, =1500 turns, core core rr=75=75

22ndnd: dia. D: dia. D22=3 cm, N=3 cm, N22=1200 turns, =1200 turns, outside 1outside 1stst

FindFind:: a) La) L11=? for the inner solenoid=? for the inner solenoid

2 220 1 11 1 1

1

7 2 2

1

4

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

4(.50m)

.133 H = 133 mH

r N DN SL

l l

L


►Exercise 3 (continued)Exercise 3 (continued) FindFind: : b) Lb) L22 = ? for the outer solenoid = ? for the outer solenoid Note: this solenoid has inner core and Note: this solenoid has inner core and

outer air outer air filled regions as in Exercise filled regions as in Exercise 1 part c), so 1 part c), so it may be treated it may be treated the same way!the same way!

2 0.087 H 87 mHL


►Exercise 3 (continued)Exercise 3 (continued) FindFind: M = ? between the two solenoids: 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 SM M M

I l

M

M

L L

individual solenoid)

SummarySummary

► Inductance results from magnetic flux Inductance results from magnetic flux ((mm) generated by electric current in a ) generated by electric current in a conductorconductor Self-inductance (L) occurs if it links with itselfSelf-inductance (L) occurs if it links with itself Mutual inductance (M) occurs if it links with Mutual inductance (M) occurs if it links with

another separate conductoranother separate conductor► The amount of inductance depends onThe amount of inductance depends on

How much magnetic flux links How much magnetic flux links How many loops the flux passes throughHow many loops the flux passes through The amount of current that generated the fluxThe amount of current that generated the flux

SummarySummary

► Inductance formulas may be derived fromInductance formulas may be derived from Direct application of the definitionDirect application of the definition Energy approachEnergy approach Vector Potential MethodVector Potential Method

► The self-inductance of some common The self-inductance of some common structures with sufficient symmetry have an structures with sufficient symmetry have an analytical resultanalytical result Coaxial cableCoaxial cable Long straight solenoidLong straight solenoid ToroidToroid Internal Inductance of a long straight wireInternal Inductance of a long straight wire

SummarySummary

►Numerical inductance may be Numerical inductance may be evaluated byevaluated by Calculation by an analytical formula if Calculation by an analytical formula if

sufficient information is known about sufficient information is known about electric current, dimensions and electric current, dimensions and permeability of materialpermeability of material

Approximation based on a curvilinear Approximation based on a curvilinear square method if axial symmetry exists square method if axial symmetry exists (uniform cross section) and a magnetic (uniform cross section) and a magnetic field map is drawn field map is drawn

ReferencesReferences

►Hayt & Buck, Hayt & Buck, Engineering Engineering ElectromagneticsElectromagnetics, 7/e, McGraw Hill: , 7/e, McGraw Hill: Bangkok, 2006.Bangkok, 2006.

►Kraus & Fleisch, Kraus & Fleisch, Electromagnetics with Electromagnetics with ApplicationsApplications, 5/e, McGraw Hill: , 5/e, McGraw Hill: Bangkok, 1999.Bangkok, 1999.

►Wentworth, Fundamentals of Wentworth, Fundamentals of Electromagnetics with Engineering Electromagnetics with Engineering Applications, John Wiley & Sons, 2005.Applications, John Wiley & Sons, 2005.

Documents

Electromagnetics ENGR 367 Inductance. Introduction Question: What physical parameters determine how much inductance a conductor or component will have