Semiconductor Electronics Materials Devices And Simple Circuits Ques 71
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 ampl, 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 :
[2016]
(a) $4,3.84$
(b) $3.69,3.84$
(c) $4,4$
(d) $4,3.69$
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Answer:
Correct Answer: 71.(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
$A_v=\alpha, \frac{R_L}{R_P}=0.96 \times \frac{800}{192}=3.333$
Power gain for common base configuration
$P_v=A_v \alpha=4 \times 0.96=3.84$
Transistor provides good power amplification when it is used in $CE$ configuration. $CC$ configuration amplifier is a current booster or voltage follower.
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