Chapter2_Notes

Impedance is given by 60 6 2 6et z0 ln s e with t

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Unformatted text preview: aterials. 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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