The shortest open-circuited length of the transmission line that appears as a capacitor of 5 pF at 2.5 GHz is approximately 0.0827 meters. The inductance of the line for an inductance of 5 nH is approximately 4.91 nH.
Step-by-step explanation:
To find the shortest open-circuited length of the transmission line that appears as a capacitor of 5 pF at 2.5 GHz, we can use the formula:
C = (ε₀εᵣL)/open-length
Given that the effective dielectric constant (εᵣ) is 1.65 and the capacitance (C) is 5 pF, we can rearrange the formula to solve for the open-length:
open-length = (ε₀εᵣL)/C
Plugging in the values, the open-length will be:
open-length = (8.854 × 10⁻¹² × 1.65 × (2.5 × 10⁹)⁻² × 5 × 10⁻¹²)/5 × 10⁻¹²
Simplifying this equation gives us:
open-length ≈ 0.0827 meters
For the inductance of 5 nH, we can use the formula:
L = (open-length)² / (ε₀εᵣC)
Using the same values as before, the inductance will be:
L = (0.0827)² / (8.854 × 10⁻¹² × 1.65 × (5 × 10⁻¹²)⁻¹²)
Simplifying this equation gives us:
L ≈ 4.91 nH