EEE340 Lecture 26 1

Internal inductance versus external inductanceExample. Magnetic-Head Connector (MHC)

Configuration of high speed magnetic head connector.

Simulation only demonstrates current distribution in ground

plane and in the two circular wires at 1 GHz.

Courteously performed under financial support of W.L. Gore

EEE340 Lecture 26 2

Current distribution in the ground plane

EEE340 Lecture 26 3

Magnetic Head Connector - 3

Fig. 6: Current distribution in the left and right circular wire

EEE340 Lecture 26 4

Measurement and simulation Results for MHC

Measurements from 1 MHz to 1 GHz

Fig. 9: Self and mutual resistances, R11, R12 by different methods

EEE340 Lecture 26 5

Measurement and Simulation Results

Self and mutual resistances, L11, L12 by different methodsQuasi-static error may exceed 300%!!

EEE340 Lecture 26 6

6-12: Magnetic Energy

Consider a single closed loop of inductance L1 without initial current.

The work done by the field (to establish magnetic fields) is

W1 is the stored magnetic energy.

11

21

11111

2

1

2

1

L

IL

dtiLdtiVWI

o

(6.157)

(6.158)

EEE340 Lecture 26 7

For a two-loop system, V21

Similarly,

The total energy

2121

I

o

212112121 IILdiILdtiVW1

(6.159)

22222 IL

2

1W

2

1

2221

121121

2

1

2

12

2

1

2

1

I

I

LL

LLII

IILWj k

kjjk

(6.161)

(6.160)

EEE340 Lecture 26 8

Generalizing the result to a system of n loops carrying currents I1, I2, …, In

where

n

2

1

nn2n1n

n22221

n11211

n21mag

I

I

I

LLL

LLL

LLL

I,,I,I2

1W

(6.162)

n

1kkkI2

1(6.166)

N

1jjjkk IL

EEE340 Lecture 26 9

6-12.1: Magnetic energy in terms of fieldsUsing

The magnetic energy (6.166) becomes

where

As

kk C

k

S

kk 'dASdB

(6.167)

kC

k

N

1kkm 'dAI

2

1W

'vJ'd)a(J'dI kkkkk

n

'V

m 'dvJA2

1W

(6.169)

EEE340 Lecture 26 10

From vector identities, we can obtain

where the integrand is magnetic energy density,

From the stored magnetic energy one can evaluate the

Inductance of a system/circuit:

'2

1'

2

1

'2

1

'

2

'

2

'

dvB

dvH

dvBHW

VV

V

m

(6-172)

2

BH

2

1HB

2

1w

22

m

(6-174)

2m

I

W2L (6-175)

EEE340 Lecture 26 11

Example 6-18 Mutual inductance. Two coils of N1 and N2

terns are wound on a cylinder core of radius a and

permeability The windings are of lengths

Find the mutual inductance.

Solution.

The magnetic flux

l2

l1

.0 21 and

a

b

11

100

21

1

1012

B

3-6 Example from where,

IN

nI

aIN

BS

EEE340 Lecture 26 12

Since coil 2 has N2 turns, we have the linkage

The magnetic coupling coefficient

The best coupling is k=1 of no leakage flux (a=b, l1=l2)

)(L:M

inductance mutual theHence,

221

1

0

1

1212

21

1

21012212

HaNNI

aINN

N

(6-151)

2

2

22

022

1

21

01

21

12

L,L

where

:

bN

aN

LL

Lk

(6-135)