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Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 36 • The various line parameters are:
R = 0 (σc = ∞) (66) G = 0 (σ = 0)
√
eﬀ
C=
Z0 c (67) 2
L = Z0 C (69) α = 0 (R = G = 0)
ω√
β=
eﬀ
c (70) (68) (71) • The expressions given allow us to determine Z0 if we are given
the dimensions and material of the microstrip transmission line
(i.e, r , h and w). Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 37 • But, to design a microstrip for a desired Z0 is more diﬃcult. So,
a family of curves are generated so that s can be estimated from
a given Z0 . We get two expressions for diﬀerent Z0 regions and
assume r is given (typical values range from 2 to 15),
• For Z0 ≤ (44 − 2 r ) Ω, w
(72)
h
2
0.52
r −1
s=
(q − 1) − ln(2q − 1) +
ln(q − 1) + 0.29 −
π
2r
r
(73)
with
60π 2
q=
(74)
√
Z0 r
s= Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 38 • For Z0 ≥ (44 − 2 r ) Ω,
s= 8ep
−2 (75) e2p with,
p= r + 1 Z0
+
2 60 r
r −1
+1 0.23 + 0.12
r Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (76) Electromagnetics I: Transmission lines 39 Microstrip Line Example: What is the width of the copper strip
for a microstrip line with Z0 = 50 Ω, and 0.5 mm thick sapphire
substrate with r = 9?
Solution: 44 − 2 r = 44 − 2 × 9 = 44 − 18 = 26 Ω (note, typo in
book), so Z0 = 50 is greater than this so we use,
p= r + 1 Z0
+
2 60 −1
r +1
r 0.23 + 0.12 (77) r when we plug in the numbers we get p = 2.06. We then get
8ep
w
= 2p
= 1.056
(78)
h
e −2
So, w = sh = 1.056 × 0.5 mm = 0.53 mm. We can check by using
s = 1.056 and r = 9 and plug into
s= 6 + (2π − 6)e−1
60
ln
+
Z0 = √
s
eﬀ 1+ 4
s2 (79) along with the other necessary equations and this gives Z0 = 49.93 Ω.
Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. Electromagnetics I: Transmission lines 40 1.6. Lossless transmission line
The most general case is a bit too complicated, so let’s make some
(very good) approximations:
• Use a conductor (wires) with low resistance; this minimizes
ohmic losses so that R
ωL .
• Use a very good dielectric between the conductors, so that G
ωC
• R ≈ 0 and G ≈ 0 results in (lossless case α = 0):
√
γ = α + jβ = 0 + jω L C
√
• And β = ω L C Notes based on Fundamentals of Applied Electromagnetics (Ulaby et al) for ECE331, PSU. (80) Electromagnetics I: Transmission lines 41 • This signiﬁcantly simpliﬁes the expression for characteristic
impedance
R + jωL
L
=
(Ω)
(81)
Z0 =
G + jωC
C
L
(Ω)
C Z0 = (82) • What do you notice about the simpliﬁed equation relative to
the original?
• Other quantities can now also be expressed in a simple form
λ= 2π
ω
1
2π
=√
, up = = √
β
...
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This note was uploaded on 09/25/2013 for the course ECE 331 taught by Professor Martinsiderious during the Fall '12 term at Portland State.
 Fall '12
 MartinSiderious
 Electromagnet

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