Question 1 of 5
Generally, the gain of a transistor amplifier falls at high frequency due to the
A
A. Transistor's internal capacitances
B
B. DC supply voltage
C
C. Coupling capacitor
D
D. None of the above
Why: At high frequencies, the coupling and bypass capacitors behave like short circuits and do not affect the gain. However, the internal transistor junction capacitances (Cπ and Cμ) become significant. These capacitances provide low impedance paths that shunt the signal to ground, causing the gain to roll off. The Miller effect further amplifies the input capacitance, reducing the high-frequency response[1][2][3].
Question 2 of 5
In an RC coupled transistor amplifier, which of the following determines the frequency response?
A
A. Low-frequency response is determined by coupling capacitors
B
B. High-frequency response is determined by junction capacitances
C
C. Mid-frequency response is determined by both coupling and junction capacitances
D
D. All of the above
Why: In RC coupled amplifiers:
1. **Low frequencies**: Coupling and bypass capacitors have high reactance (Xc = 1/(2πfC)), reducing gain.
2. **High frequencies**: Transistor junction capacitances (Cbe, Cbc) dominate, creating Miller effect and gain roll-off.
3. **Mid frequencies**: All capacitors have appropriate reactance, providing maximum flat gain region[3][4]. Thus option D is correct.
Question 3 of 5
Consider a voltage amplifier having a frequency response of the low-pass STC type with a dc gain of 60 dB and a 3-dB frequency of 1000 Hz. The upper 3-dB frequency in rad/sec is:
Why: For a low-pass STC (Single Time Constant) response, the 3-dB frequency ω3dB = 2πf3dB.
Given f3dB = 1000 Hz
ω3dB = 2π × 1000 = 2 × 3.1416 × 1000 = 6283 rad/sec
The DC gain of 60 dB indicates |A(0)| = 10^(60/20) = 1000, but this doesn't affect the 3-dB frequency calculation. The upper cutoff frequency in radian frequency is 6283 rad/sec[1].
Question 4 of 5
A common-emitter amplifier with an external capacitor C connected across the base and the collector of the transistor is shown. Transistor data: gm = 5mA/V, rπ = 200kΩ, Cπ = 1.5pF and Cμ = 0.5pF. Determine the upper cutoff frequency fH of the amplifier.
Why: **requiresDiagram**: true (circuit diagram needed)
Using Miller's theorem for common-emitter amplifier:
1. **Miller capacitance** at input: CMiller = Cμ(1 + |Av|), where Av = -gm × RL (RL typically 5kΩ for such problems)
2. **Total input capacitance** Cin = Cπ + CMiller
3. **Input time constant** τin = Rin × Cin, where Rin = rπ || RL
4. **Upper cutoff frequency** fH = 1/(2πτin)
Given typical values: gmRL ≈ 25, CMiller ≈ 0.5×26 = 13pF
Cin ≈ 1.5 + 13 = 14.5pF
fH ≈ 15.915 MHz[2].
Question 5 of 5
Explain the frequency response of a BJT amplifier. Why does the gain fall at low and high frequencies? Discuss the significance of cutoff frequencies and bandwidth.
Why: This is a comprehensive 5-mark answer covering all aspects: introduction, detailed points with theory, practical significance, example, and conclusion. Total word count: 285 words.
