Math Problem Statement

Calculate what L network components will match a 42 internal resistance in series with an internal inductance of 9 nH to a 72-2 load impedance in parallel with a stray capacitance of 24 pF at a frequency of 46 MHz.

Solution

To design an L-network that matches a source impedance of Zs=42+jωLsZ_s = 42 + j\omega L_s (where Ls=9 nHL_s = 9 \text{ nH}) to a load impedance of ZL=72jωCpZ_L = 72 - j\omega C_p (where Cp=24 pFC_p = 24 \text{ pF}), we can follow these steps:

  1. Calculate the Impedances of the Source and Load at 46 MHz:

    • Source Impedance: Zs=42+jωLsZ_s = 42 + j\omega L_s

      • Here, ω=2π46×1062.89×108rad/s\omega = 2\pi \cdot 46 \times 10^6 \approx 2.89 \times 10^8 \, \text{rad/s}.
      • The inductive reactance, Xs=ωLs=(2.89×108)9×109=2.6ΩX_s = \omega L_s = (2.89 \times 10^8) \cdot 9 \times 10^{-9} = 2.6 \, \Omega.
      • So, Zs=42+j2.6ΩZ_s = 42 + j2.6 \, \Omega.
    • Load Impedance: ZL=72j1ωCpZ_L = 72 - j\frac{1}{\omega C_p}

      • The capacitive reactance, Xp=1ωCp=1(2.89×108)24×1012=14.4ΩX_p = \frac{1}{\omega C_p} = \frac{1}{(2.89 \times 10^8) \cdot 24 \times 10^{-12}} = 14.4 \, \Omega.
      • Thus, ZL=72j14.4ΩZ_L = 72 - j14.4 \, \Omega.
  2. Choosing L-Network Configuration: Given that the source has a real part lower than the load's real part, a "high-pass" L-network configuration (series inductance and parallel capacitance) is suitable.

  3. Calculate the Required Components:

    • Calculate the Q-factor (Q): Q=RLRs1=724210.85Q = \sqrt{\frac{R_L}{R_s} - 1} = \sqrt{\frac{72}{42} - 1} \approx 0.85

    • Determine the Series Inductor (L): Xs=QRs=0.854235.7ΩX_s = Q \cdot R_s = 0.85 \cdot 42 \approx 35.7 \, \Omega L=Xsω=35.72.89×108123.5nHL = \frac{X_s}{\omega} = \frac{35.7}{2.89 \times 10^8} \approx 123.5 \, \text{nH}

    • Determine the Parallel Capacitor (C): Xp=RLQ=720.8584.7ΩX_p = \frac{R_L}{Q} = \frac{72}{0.85} \approx 84.7 \, \Omega C=1ωXp=1(2.89×108)84.741.1pFC = \frac{1}{\omega X_p} = \frac{1}{(2.89 \times 10^8) \cdot 84.7} \approx 41.1 \, \text{pF}

Summary of Component Values:

  • Series Inductor (L): 123.5nH123.5 \, \text{nH}
  • Parallel Capacitor (C): 41.1pF41.1 \, \text{pF}

This L-network will match the given source and load impedances at the operating frequency of 46 MHz.

Would you like a detailed explanation of each step or any clarifications?


  1. How do L-networks differ from Pi-networks in impedance matching?
  2. What happens if the Q-factor calculated is very low or very high?
  3. How can you determine if a high-pass or low-pass L-network configuration is needed?
  4. Why does the operating frequency affect the L and C values in the network?
  5. How do stray capacitances affect the matching network at high frequencies?

Tip: When designing matching networks, always verify the operating frequency and ensure that component tolerances align with design requirements for accuracy.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Impedance Matching
Complex Impedance
Reactance Calculations
L-Network Design

Formulas

Impedance of inductor: Z = jωL
Impedance of capacitor: Z = -j / ωC
Q-factor: Q = √(R_L/R_s - 1)
Series inductance: X_s = Q * R_s
Parallel capacitance: X_p = R_L / Q

Theorems

Impedance Matching for L-Networks
Reactive Power Compensation

Suitable Grade Level

Undergraduate Engineering