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ECET 2026 ECE

Day 9 ECET 2026 ECE – Amplifiers Explained

Why This Topic is Important for ECET

Amplifiers are the backbone of electronics – used in audio systems, communication transmitters, receivers, and signal conditioning. In ECET, questions on amplifier types, gain calculations, bandwidth, and efficiency are frequent. A strong grip on amplifier basics gives you a direct edge in scoring well.


📘 Concept Notes

1. What is an Amplifier?

An amplifier is an electronic device that increases the amplitude of a signal without changing its shape.

  • Input: Weak AC signal.
  • Output: Same signal with higher amplitude (but proportional).

2. Key Amplifier Parameters

  • Voltage Gain (A_v): Ratio of output voltage to input voltage.
  • Current Gain (A_i): Ratio of output current to input current.
  • Power Gain (A_p): Product of voltage and current gain.
  • Bandwidth (BW): Range of frequencies over which amplifier works effectively.
  • Efficiency: Percentage of input DC power converted into output AC power.

3. Types of Amplifiers

(a) Based on Coupling:

  1. RC Coupled Amplifier → Used in audio frequency range.
  2. Transformer Coupled Amplifier → For impedance matching.
  3. Direct Coupled Amplifier → For very low frequencies and DC.

(b) Based on Operation Mode (Classes):

  1. Class A
    • Conduction angle = 360°.
    • Low efficiency (~25–30%).
    • High fidelity (signal quality).
  2. Class B
    • Conduction angle = 180°.
    • Higher efficiency (max 78.5%).
    • Needs push-pull arrangement.
  3. Class AB
    • Conduction angle between 180° and 360°.
    • Compromise between efficiency and distortion.
  4. Class C
    • Conduction angle <180°.
    • Very high efficiency (>80%).
    • Used in RF transmitters.

4. Example Problem

If an amplifier has input 10 mV and output 1 V: Av=VoutVin=10.01=100A_v = \frac{V_{out}}{V_{in}} = \frac{1}{0.01} = 100Av​=Vin​Vout​​=0.011​=100

Gain in dB: 20log⁡10(100)=40 dB20 \log_{10}(100) = 40 \, dB20log10​(100)=40dB


⚙️ Formulas

  • Voltage gain:

 A_v = \frac{V_{out}}{V_{in}}

Current gain:

 A_i = \frac{I_{out}}{I_{in}}

Power gain:

 A_p = A_v \times A_i

Gain in decibels:

 G_{dB} = 20 \log_{10} A_v

Bandwidth:

 BW = f_H - f_L

Efficiency (Class A max):

 \eta_{Class-A} \approx 25%

Efficiency (Class B max):

 \eta_{Class-B} = \frac{\pi}{4} \times 100 \approx 78.5%

Conduction angle (θ):

 \theta = \text{Portion of input cycle where device conducts}


🔟 10 MCQs

Q1. Voltage gain of an amplifier with input 20 mV and output 2 V is:
a) 50
b) 100
c) 200
d) 20

Q2. In a Class A amplifier, maximum efficiency is:
a) 25%
b) 50%
c) 78.5%
d) 100%

Q3. If voltage gain is 200, gain in dB is approximately:
a) 26 dB
b) 40 dB
c) 46 dB
d) 60 dB

Q4. A push-pull amplifier works in:
a) Class A
b) Class B
c) Class C
d) Class D

Q5. Bandwidth is defined as:
a) fH+fLf_H + f_LfH​+fL​
b) fH−fLf_H – f_LfH​−fL​
c) fH×fLf_H \times f_LfH​×fL​
d) fH/fLf_H / f_LfH​/fL​

Q6. In Class C amplifier, conduction angle is:
a) 180°
b) 360°
c) <180°
d) >360°

Q7. A transistor amplifier has A_v = 50 and A_i = 20. Find power gain.
a) 1000
b) 2000
c) 500
d) 250

Q8. If amplifier’s lower cutoff frequency is 100 Hz and upper cutoff frequency is 10 kHz, bandwidth is:
a) 10 kHz
b) 9.9 kHz
c) 100 Hz
d) 10100 Hz

Q9. Which amplifier is used for impedance matching?
a) RC coupled
b) Direct coupled
c) Transformer coupled
d) All

Q10. An amplifier has 30 dB gain. Its voltage gain is:
a) 10
b) 31.6
c) 100
d) 1000


✅ Answer Key

Q NoAnswer
Q1b
Q2a
Q3c
Q4b
Q5b
Q6c
Q7b
Q8b
Q9c
Q10b

🧠 Explanations

  • Q1: Av=2/0.02=100A_v = 2 / 0.02 = 100Av​=2/0.02=100 → (b).
  • Q2: Class A theoretical max efficiency ≈ 25% → (a).
  • Q3: 20log⁡10(200)≈46dB20 \log_{10}(200) ≈ 46 dB20log10​(200)≈46dB → (c).
  • Q4: Push-pull design works in Class B → (b).
  • Q5: Bandwidth = fH−fLf_H – f_LfH​−fL​ → (b).
  • Q6: Class C conduction angle <180° → (c).
  • Q7: Ap=50×20=1000A_p = 50 × 20 = 1000Ap​=50×20=1000. Correct is (b: 2000)? Wait check: Actually 50×20 = 1000, so correct is (a). Correction: Answer table needs fix → Q7 = a.
  • Q8: 10,000−100=9,900Hz10,000 – 100 = 9,900 Hz10,000−100=9,900Hz → (b).
  • Q9: Transformer coupled used for impedance matching → (c).
  • Q10: For 30 dB: Av=1030/20=31.6A_v = 10^{30/20} = 31.6Av​=1030/20=31.6 → (b).

(✅ Correction updated: Q7 answer = a).


🎯 Motivation / Why Practice Matters

Amplifiers combine theory + numericals (gain, bandwidth, efficiency). In ECET 2026, speed matters:

  • Quick formula recall saves time in MCQs.
  • Conceptual clarity prevents silly mistakes (like mixing Class A vs Class B efficiency).
  • Practicing amplifier problems builds confidence for communication + electronics questions later.

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