Semiconductor Electronics Materials Devices and Simple Circuits - Result Question 61
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63. A npn transistor is connected in common emitter configuration in a given amplifier. A load resistance of $800 \Omega$ is connected in the collector circuit and the voltage drop across it is $0.8 V$. If the current amplification factor is 0.96 and the input resistance of the circuit is $192 \Omega$, the voltage gain and the power gain of the amplifier will respectively be :
======= ####63. A npn transistor is connected in common emitter configuration in a given amplifier. A load resistance of $800 \Omega$ is connected in the collector circuit and the voltage drop across it is $0.8 V$. If the current amplification factor is 0.96 and the input resistance of the circuit is $192 \Omega$, the voltage gain and the power gain of the amplifier will respectively be :
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/semiconductor-electronics-materials-devices-and-simple-circuits/semiconductor-electronics-materials-devices-and-simple-circuits—result-question-61.md (a) $4,3.84$
(c) 4,4
(b) $3.69,3.84$
(d) $4,3.69$
[2016]
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Answer:
Correct Answer: 63. (a)
Solution:
- (a) Given: amplification factor $\alpha=0.96$
load resistance, $R_L=800 \Omega$
input resistance, $R_i=192 \Omega$
So, $\beta=\frac{\alpha}{1-\alpha}=\frac{0.96}{0.04} \Rightarrow \beta=24$
Voltage gain for common emitter configuration
$A_v=\beta \cdot \frac{R_L}{R_i}=24 \times \frac{800}{192}=100$
Power gain for common emitter configuration
$P_v=\beta A_v=24 \times 100=2400$
Voltage gain for common base configuration
$A_v=\alpha, \frac{R_L}{R_P}=0.96 \times \frac{800}{192}=4$
Power gain for common base configuration
$P_v=A_v \alpha=4 \times 0.96=3.84$
Transistor provides good power amplification when they are used in CE configuration. CC configuration amplifier is a power amplifier or current booster.
