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Internal inductance versus external inductance

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Presentation on theme: "Internal inductance versus external inductance"— Presentation transcript:

1 Internal inductance versus external inductance
Example. 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 methods Quasi-static error may exceed 300%!! EEE340 Lecture 26

6 The work done by the field (to establish magnetic fields) is
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. (6.157) (6.158) EEE340 Lecture 26

7 For a two-loop system, V21 Similarly, The total energy (6.159) (6.160)
(6.161) EEE340 Lecture 26

8 Generalizing the result to a system of n loops carrying currents I1, I2, …, In
where (6.162) (6.166) EEE340 Lecture 26

9 6-12.1: Magnetic energy in terms of fields
Using The magnetic energy (6.166) becomes where As (6.167) (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: (6-172) (6-174) (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 b a 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) (6-151) (6-135) EEE340 Lecture 26

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