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Tuned amplifier applications, bandwidth, Q factor

Short answer

Used in RF/IF amplifiers, transmitters, and narrow‑band systems. Resonance frequency fr = 1/(2π√(LC)); quality factor Q = ωr L/r; 3 dB bandwidth BW = fr/Q. The tank’s loaded Q determines selectivity, and gain peaks at fr.

What are the applications of tuned amplifiers? Determine its 3 dB Bandwidth, Resonance frequency and Quality factor.

Applications of tuned amplifiers

  • RF amplifiers at the front end of radio and TV receivers (select one station and reject others).
  • IF amplifiers in superheterodyne receivers (for example 455 kHz in AM, 10.7 MHz in FM).
  • Transmitters: tuned power amplifiers and frequency multipliers.
  • Communication systems where a narrow band must be amplified: wireless links, radar, telemetry.
  • Band-pass filtering together with amplification.

Resonance frequency, Q and 3 dB bandwidth

A tuned amplifier uses a parallel LC tank (coil inductance L with series resistance r, and capacitor C) as the collector or drain load.

Resonance frequency:

fr = 1 / (2π √(LC)) (for Q > 10)

At resonance the tank impedance is purely resistive and maximum: Zr = Rp = L/(C r) = Q ωr L.

Quality factor:

Q = ωr L / r = Rp/(ωr L)

When the transistor output resistance and the load are in parallel with the tank, Rp is replaced by the total effective resistance Reff, giving the loaded Q: QL = Reff/(ωr L).

3 dB bandwidth:

BW = f2 − f1 = fr/Q

where f1 and f2 are the frequencies at which the gain falls to 0.707 of the maximum:

f1 ≈ fr − BW/2, f2 ≈ fr + BW/2

Voltage gain at resonance: Av = −gm Reff (maximum). At other frequencies, Av = −gm Reff / (1 + j Q (f/fr − fr/f)) ≈ −gm Reff / (1 + j 2Q δ), with δ = (f − fr)/fr.

Example

L = 100 µH, C = 253 pF, loaded Q = 50:

fr = 1/(2π√(100×10⁻⁶ × 253×10⁻¹²)) = 1.0 MHz, BW = 1 MHz/50 = 20 kHz, so f1 = 990 kHz and f2 = 1010 kHz.

A higher Q gives a narrower band and better selectivity.

✓ Answer checked by Claude Opus 5.5 (high effort)

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