- Asked 16/28
- 2083 Baisakh · 6 marks
- 2082 Shrawan · 4+3 marks
- 2081 Baisakh · 3+3 marks
- 2080 Bhadra · 6+2 marks
- 2078 Bhadra · 2+3 marks
- 2078 Kartik · 7 marks
- 2076 Chaitra · 5 marks
- 2075 Chaitra · 6 marks
- 2074 Chaitra · 6 marks
- 2074 Ashwin · 6 marks
- 2073 Shrawan · 3+3 marks
- 2072 Chaitra · 1+1+1+1 marks
- 2068 Chaitra · 2+6 marks
- 2072 Kartik · 6 marks
- 2071 Shrawan · 5 marks
- 2070 Chaitra · 2+5 marks
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)
Related questions
- Series fed Class A amplifier maximum efficiency · Electronics Devices and Circuit, chapter 4 · asked 9×
- Tuned amplifier applications, bandwidth, Q factor · Electronics Devices and Circuit, chapter 4
- Synchronous vs stagger tuned amplifier · Electronics Devices and Circuit, chapter 4
- Design class B amplifier 20 W into 8 Ω · Electronics Devices and Circuit, chapter 4
- Transformer coupled class B: 78.5% and maximum loss · Electronics Devices and Circuit, chapter 4
- Class B power amplifier efficiency · Electronics Devices and Circuit, chapter 4 · asked 8×
- GaAs FET amplifier for maximum gain at 4 GHz · RF and Microwave Engineering, chapter 6
- Slew-rate check for 40 dB amplifier at 20 kHz · Advanced Electronics, chapter 2
- R-H range of k and oscillation frequency · Control Systems, chapter 4 · asked 2×
- Active load doubles differential amplifier gain · Advanced Electronics, chapter 1 · asked 2×