Tuned Amplifiers: amplify one frequency, reject the rest
Replace the collector resistor with an LC tank and you've built the heart of a radio.
An audio amplifier should treat all frequencies equally. A radio amplifier must do the opposite: pick out one station at 1000 kHz and reject the one at 1010 kHz. The change needed is small: swap the collector load resistor for a parallel LC tank. But the consequences are large. Why? The tank's impedance is huge at resonance and small everywhere else, so the stage amplifies only near that one frequency.
The principle
Gain of a CE stage: A_v = −g_m × Z_load
With a resistor: Z_load = R_C, constant at all frequencies
→ flat response
With an LC tank: Z_load = maximum at f₀, low elsewhere
→ gain PEAKS sharply at f₀
At resonance the tank's dynamic impedance is
Z_max = L/(C·R) = Q × ω₀L = Q²R
so the gain at f₀ is high, and falls away on both sides.
The key parameters
Resonant frequency: f₀ = 1/(2π√(LC))
Quality factor: Q = ω₀L/R = (1/R)√(L/C)
Bandwidth: BW = f₀/Q
Gain at resonance: A_v = −g_m × Z_max = −g_m Q ω₀ L
Selectivity ∝ Q. Higher Q → narrower BW → better at
rejecting adjacent channels, but distorts wideband signals.
There is an unavoidable trade between selectivity and bandwidth, and it's set entirely by Q. A radio's IF strip needs enough bandwidth to pass the modulation (say 10 kHz for AM audio) but no more, so adjacent stations are rejected. Choosing Q is therefore choosing the bandwidth: BW = f₀/Q, with no way around it in a single-tuned stage.
Worked numerical 1 — full tuned amplifier analysis
A tuned amplifier has L = 150 µH, C = 220 pF, coil resistance R = 12 Ω, and the transistor has g_m = 40 mS. Find f₀, Q, bandwidth, tank impedance and voltage gain.
Resonant frequency:
f₀ = 1/(2π√(LC))
= 1/(2π√(150 × 10⁻⁶ × 220 × 10⁻¹²))
= 1/(2π√(3.3 × 10⁻¹⁴))
= 1/(2π × 1.8166 × 10⁻⁷)
= 876.3 kHz (in the AM broadcast band)
ω₀ = 2π × 876 300 = 5.507 × 10⁶ rad/s
Quality factor:
Q = ω₀L/R = (5.507 × 10⁶ × 150 × 10⁻⁶)/12
= 826/12 = 68.8
Bandwidth:
BW = f₀/Q = 876 300/68.8 = 12.74 kHz
→ passes an AM channel (10 kHz) nicely ✔
Tank dynamic impedance at resonance:
Z_max = L/(CR) = (150 × 10⁻⁶)/(220 × 10⁻¹² × 12)
= 1.5 × 10⁻⁴/2.64 × 10⁻⁹
= 56 818 Ω ≈ 56.8 kΩ
(check via Q²R = 68.8² × 12 = 56 800 Ω ✔)
Voltage gain at resonance:
A_v = −g_m × Z_max = −0.040 × 56 818
= −2273
Off resonance, say at 800 kHz, the tank impedance collapses
to a few hundred ohms and the gain falls to single digits —
that difference IS the selectivity.
Worked numerical 2 — designing for a required bandwidth
Design a tuned stage for f₀ = 455 kHz (the standard AM intermediate frequency) with a 10 kHz bandwidth, using C = 500 pF.
Required Q:
Q = f₀/BW = 455 000/10 000 = 45.5
Inductance from the resonance condition:
L = 1/(4π²f₀²C)
= 1/(4π² × (4.55 × 10⁵)² × 500 × 10⁻¹²)
= 1/(39.478 × 2.070 × 10¹¹ × 5 × 10⁻¹⁰)
= 1/(4086)
= 2.447 × 10⁻⁴ H = 244.7 µH
Required coil resistance for Q = 45.5:
Q = ω₀L/R → R = ω₀L/Q
ω₀ = 2π × 455 000 = 2.859 × 10⁶
R = (2.859 × 10⁶ × 244.7 × 10⁻⁶)/45.5
= 699.6/45.5 = 15.4 Ω
If the actual coil has only 8 Ω, Q would be 87 and BW only
5.2 kHz — too narrow, cutting the audio. The fix is to ADD
a damping resistor in parallel with the tank to deliberately
lower Q. Widening bandwidth by throwing away gain is a
standard technique.
Worked numerical 3 — cascaded stages narrow the bandwidth
Three identical tuned stages, each with 12 kHz bandwidth, are cascaded. Find the overall bandwidth.
For n identical cascaded single-tuned stages:
BW_total = BW_single × √(2^(1/n) − 1)
For n = 3:
2^(1/3) = 1.2599
√(1.2599 − 1) = √0.2599 = 0.5098
BW_total = 12 × 0.5098 = 6.12 kHz
The bandwidth SHRINKS to about half — cascading for more
gain costs bandwidth. Three stages of 12 kHz give 6.1 kHz,
which would now clip the AM audio.
The remedy is STAGGER TUNING: set the three stages to
450, 455 and 460 kHz. Individually each is narrow, but their
overlapping responses sum to a wide, flat-topped passband —
exactly what a receiver's IF strip needs.
💡 Exam angle: the reliable numerical is f₀, Q, BW and Z_max = L/CR — memorise that Z_max also equals Q²R, which gives a free cross-check. For descriptive marks, explain the selectivity/bandwidth trade-off (BW = f₀/Q) and name the three types (single-, double-, stagger-tuned). The cascading formula BW√(2^(1/n) − 1) is a bonus but occasionally asked.
Syllabus points
Tuned amplifier operation
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