EEE340Lecture 261 Internal inductance versus external inductance Example. Magnetic-Head Connector...
-
date post
21-Dec-2015 -
Category
Documents
-
view
236 -
download
0
Transcript of EEE340Lecture 261 Internal inductance versus external inductance Example. Magnetic-Head Connector...
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 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