--- title: "Class A tuned amplifier frequency range" subject: "Electronics Devices and Circuit" chapter: 4 key: "semester_3/electronics_devices_and_circuit:tuned-amplifier-3db-bandwidth" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/tuned-amplifier-3db-bandwidth site: BCT Engineering --- # Class A tuned amplifier frequency range Electronics Devices and Circuit · Chapter 4: Output Stages and Power Amplifiers - Asked in: 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 (asked 16 times) - Video: [TUNED AMPLIFIER](https://www.youtube.com/watch?v=eVa7GOzGZ1s) — TutRoNiCS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/tuned-amplifier-3db-bandwidth - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-4 ## Question 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. ## 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. ## Answer 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 ```text +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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the collector load of a class A tuned amplifier? A. A parallel LC tank circuit B. A plain resistor C. A transformer only D. A diode Answer: A. The parallel LC tank gives a high impedance only near resonance. (Key point: Tuned load) 2. The resonant frequency is: A. 1/(2π sqrt(LC)) B. 2π sqrt(LC) C. 1/(LC) D. sqrt(L/C) Answer: A. fr = 1/(2π sqrt(LC)), where the tank impedance is maximum and resistive. (Key point: Resonant frequency) 3. Within 3 dB of maximum gain, the frequency range is: A. fr ± fr/(2Q) B. fr ± Q C. fr ± 2Q fr D. 0 to fr Answer: A. BW = fr/Q, and f1,f2 = fr ∓ BW/2. (Key point: 3 dB range) 4. What does a higher loaded Q do? A. Narrows bandwidth, more selectivity B. Widens bandwidth C. Shifts fr D. Increases gain at all frequencies Answer: A. BW = fr/Q, so larger Q gives a narrower, more selective band (e.g. 20 kHz at Q = 50, fr = 1 MHz). 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