Digital Signal Processing Reference
In-Depth Information
similar to loop (a). As frequency increases, the skin effect will begin to force
the current to flow on the periphery of the conductor, so internal currents will
decrease. At very high frequencies, there will be almost no current flowing in
loop (a) because the skin effect has forced all the current to flow on the sur-
face, causing the inductance of loop (b) to dominate. The inductance associated
with loop (a), where the currents are flowing internal to the conductor, is com-
monly known as the internal inductance . The inductance associated with loop
(b), where all currents are flowing on the surface of the conductor, is commonly
known as the external inductance . The total inductance is the sum of the external
and internal parts:
L total
= L internal
+ L external
(5-20)
As frequency increases, the skin effect causes the current to flow very close to
the surface, with minimal current flowing interior to the conductor. At infinite
frequencies, all the current is flowing on the surface and the internal contribution
is zero. Consequently, as frequency increases, the effect of L internal will decrease
and L total will asymptote to L external .
The external inductance is calculated with quasistatic techniques assuming that
all the charge is on the surface of the conductor and solving Laplace's equation
as shown in Sections 3.4.2, 3.4.3, and 3.4.6. The internal inductance is derived
by observing Ampere's (2-34) and Faraday's (2-33) laws for a good conductor.
For a good conductor , the conduction current
J will be much larger than the
displacement current jω D , and therefore Ampere's law reduces to
∇× H
J
= σ E
(5-21)
Faraday's law is repeated here:
∇× E + jωµ H
=
0
(2-33)
Taking the curl of (2-33) yields the following equation:
∇× ( ∇× E) =− jωµ ∇× H
(5-22)
If it is assumed that the free charge is negligible and the source of the electric field
is the time-varying magnetic field, Gauss's law reduces to
∇· ε E =
0, allowing
the use of the vector identity (see Appendix A)
∇× ( ∇× ψ) =∇ ( ∇· ψ) −∇
2
ψ
∇× ( ∇× E) =
E =− jωµ ∇× H
2
0
−∇
to simplify the equation, which reduces to
2
E = jωµ ∇× H
(5-23)
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