Parallel resonant band-pass design: A 6.8 Ω series resistor feeds a parallel tank of L = 8 µH and C = 120 pF (RW ≈ 0 Ω). The output is taken across the LC. What is the center frequency f0?

Difficulty: Medium

Correct Answer: 5.14 MHz

Explanation:


Introduction / Context:
This item tests the core resonance relation for a parallel LC tank, commonly used in RF band-pass filters. The series source resistor sets loading, but the ideal center frequency is set by L and C alone when winding resistance is negligible.


Given Data / Assumptions:

  • Inductance L = 8 µH.
  • Capacitance C = 120 pF.
  • Series feed resistor = 6.8 Ω (does not change f0 ideally).
  • Inductor winding resistance RW = 0 Ω (assumed).


Concept / Approach:
The center (resonant) frequency for an LC tank is f0 = 1 / (2 * pi * sqrt(L * C)). The exact f0 is largely independent of a small series feed resistor; resistive elements mainly affect bandwidth and Q, not the ideal resonant frequency.


Step-by-Step Solution:

Compute LC: L * C = 8 * 10^-6 * 120 * 10^-12 = 9.6 * 10^-16.Take square root: sqrt(L * C) ≈ 3.098 * 10^-8.Apply formula: f0 = 1 / (2 * pi * 3.098 * 10^-8) ≈ 1 / (1.946 * 10^-7) ≈ 5.14 * 10^6 Hz.Thus, f0 ≈ 5.14 MHz.


Verification / Alternative check:
A quick RF check: Using the rule-of-thumb f0(MHz) ≈ 159 / sqrt(L(µH) * C(pF)) gives 159 / sqrt(8 * 120) ≈ 159 / 31.06 ≈ 5.12 MHz, very close to 5.14 MHz.


Why Other Options Are Wrong:

  • 514 kHz and 503 kHz: Off by a factor of 10 (wrong power-of-ten handling).
  • 5.03 MHz: Slightly low due to rounding; 5.14 MHz matches precise calculation.


Common Pitfalls:

  • Mistaking series vs. parallel resonance formulas.
  • Mixing µH and mH, or pF and nF, leading to decade errors.


Final Answer:
5.14 MHz

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