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TRANSCRIPT
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Inductances and
InductorsDragica Vasileska
ProfessorArizona State University
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Flux Linkage
Consider two magnetically coupled circuits
C1I1
S1 S2 C2
I2
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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
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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
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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
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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
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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
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Self Inductance (Contd)
For an isolated circuit, we call the selfinductance, inductance, and evaluate itusing
I
N
I
L
2
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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
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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
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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
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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
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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
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Energy Stored in an Inductor
The magnetic energy stored in an inductoris given by
2
2
1LIW
m
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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
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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
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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
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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
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Example of Calculating
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
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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
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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.
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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
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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
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Example of Calculating
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
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Example of Calculating
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
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Example of Calculating
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)
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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.
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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:
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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 :
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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
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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