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What is the width when Zo is 125 ohms?

What is the width when Zo is 125 ohms?

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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What is the width when Zo is 125 ohms?
I assume you meant "dielectric constant = 3.48, thickness = 0.76 -diaelectric thickness". To solve for the width, we need to rearrange the first equation to isolate the width: W = -2d e We can use the second equation to find the effective dielectric constant, which we will substitute into the first equation to solve for the characteristic impedance, z0: |5eff = £0+² < ft(1 + 12d right)¹] 2 W Substituting the given values, we get: < ft (1+ Eeff = 8d zo(√ff60 Zo - 1+3.48 2 Solving for zo using the first equation: 60 Zo = √6 In<ft (+right) = 0 In<ft (+right) √Eff √3.48 Substituting the expressions for w and 20 into the equation for the characteristic impedance of a transmission line: Zo We get: = 12d right)¹ = 2.24 ≤ ft(1 + W 120T √ Eeff 120T √2.24 12d right)¹] W +1.393+0.667ln ≤ft(+1.444right) 20√ Eff ZO√/Eff 60 2+1.393+0.667ln <fte 60 2 +1.444right /e V Substituting the given values into the second equation, we get: Eeff 1+3.48 < ft (1 + 12×0.76 right) ¹ = 2.24 < ft (1 + 9,12 right)¹] 2 W W Substituting this into the first equation, we get: 60 In ≤ft (8 + right) ≈ 93.93ln <ft (8 + right) 8d √3.48 W W ZO = Using the equation for w that we derived earlier and solving numerically, we get: [w≈ 0.74, mm] Therefore, the width of the transmission line is approximately 0.74 mm.
Transcribed Image Text:I assume you meant "dielectric constant = 3.48, thickness = 0.76 -diaelectric thickness". To solve for the width, we need to rearrange the first equation to isolate the width: W = -2d e We can use the second equation to find the effective dielectric constant, which we will substitute into the first equation to solve for the characteristic impedance, z0: |5eff = £0+² < ft(1 + 12d right)¹] 2 W Substituting the given values, we get: < ft (1+ Eeff = 8d zo(√ff60 Zo - 1+3.48 2 Solving for zo using the first equation: 60 Zo = √6 In<ft (+right) = 0 In<ft (+right) √Eff √3.48 Substituting the expressions for w and 20 into the equation for the characteristic impedance of a transmission line: Zo We get: = 12d right)¹ = 2.24 ≤ ft(1 + W 120T √ Eeff 120T √2.24 12d right)¹] W +1.393+0.667ln ≤ft(+1.444right) 20√ Eff ZO√/Eff 60 2+1.393+0.667ln <fte 60 2 +1.444right /e V Substituting the given values into the second equation, we get: Eeff 1+3.48 < ft (1 + 12×0.76 right) ¹ = 2.24 < ft (1 + 9,12 right)¹] 2 W W Substituting this into the first equation, we get: 60 In ≤ft (8 + right) ≈ 93.93ln <ft (8 + right) 8d √3.48 W W ZO = Using the equation for w that we derived earlier and solving numerically, we get: [w≈ 0.74, mm] Therefore, the width of the transmission line is approximately 0.74 mm.
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