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Class A tuned amplifier frequency range

Short answer

A class A tuned amplifier uses a parallel LC load to provide maximum gain at resonance fr=1/(2π√(LC)). The 3dB bandwidth is BW=fr/Q, so the frequency range within 3dB of maximum gain is fr ± fr/(2Q); e.g., for Q=50, BW=20kHz around 1MHz.

Draw the circuit diagram of Class A tuned amplifier and determine the range of frequency in which it gives maximum gain within 3 dB range.

A class A tuned amplifier uses a parallel LC tank circuit as the collector load. It amplifies a narrow band of frequencies around the resonant frequency and rejects the rest. It is used in radio and IF stages.

Circuit diagram

                      +VCC
                        |
         +--------------+
         |              |
         |        +-----+-----+
         |        |           |
        [R1]     [L]         [C]   tank (parallel LC)
         |        |           |
         |        +-----+-----+
         |              |
         |              +------||------o Vout
         |              |      C2
         |              C
Vin o-||-+--------------B   Q1 (NPN)
     C1  |              E
        [R2]            |
         |              +-------+
         |              |       |
        GND           [RE]    [CE] (bypass)
                        |       |
                       GND     GND

R1 (from VCC to the base) and R2 (from the base to ground) give the voltage-divider bias (class A, Q point at the middle of the load line). RE with the bypass capacitor CE gives stability. C1 and C2 are coupling capacitors.

Operation

  • Resonant frequency: fr = 1 / (2π √(LC))
  • At resonance the tank impedance is purely resistive and maximum: Rp = Q ωr L = Q² R_coil. The voltage gain is then maximum: Av(max) = −gm (Rp ∥ ro ∥ Ri(next stage)).
  • Away from fr the tank impedance falls, so the gain falls.

Frequency range within 3 dB of maximum gain

The 3 dB points are where the gain drops to 0.707 of the maximum.

  • Quality factor (loaded): Q = Rp/(ωr L) = fr/BW
  • Bandwidth: BW = f2 − f1 = fr/Q
  • Lower and upper 3 dB frequencies (for Q > 10):

f1 = fr − BW/2, f2 = fr + BW/2

So the amplifier gives maximum gain (within 3 dB) over the range f1 to f2 = fr ± fr/(2Q).

Example

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

fr = 1/(2π √(100×10⁻⁶ × 253×10⁻¹²)) = 1.0 MHz

BW = fr/Q = 1 MHz/50 = 20 kHz

f1 = 1000 − 10 = 990 kHz, f2 = 1000 + 10 = 1010 kHz

Increasing Q narrows the band (more selectivity); decreasing Q widens it.

✓ Answer checked by Claude Opus 5.5 (high effort)

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