Modern Physics Ques 250
- An $n-p-n$ transistor operates as a common emitter amplifier, with a power gain of $60 $ $dB$. The input circuit resistance is $100$ $\Omega$ and the output load resistance is $10 $ $ k \Omega$. The common emitter current gain $\beta$ is
(Main 2019, 10 April I)
(a) $10^{2}$
(b) $6 \times 10^{2}$
(c) $10^{4}$
(d) $60$
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Answer:
Correct Answer: 250.(a)
Solution:
Formula:
- Given, $A _P=60 $ $dB$ (in decibel)
Power gain in decibel can be given as
$ \begin{aligned} & A _P=10 \log _{10} \left(\frac{\text { Output power }}{\text { Input power }}\right) \\ \Rightarrow & \quad 60=10 \log _{10} \left(\frac{P _{\text {out }}}{P _{\text {in }}}\right) \Rightarrow \log _{10} (\frac{P _{\text {out }}}{P _{\text {in }}})=6 \\ \Rightarrow & \quad \frac{P _{\text {out }}}{P _{\text {in }}}=10^{6}=A _P \end{aligned} $
Also, given $R _{\text {out }}=10 $ $k \Omega, R _{\text {in }}=100$ $ \Omega$
$\therefore$ Power gain of a transistor is given by
$ A _P=\beta^{2} \left(\frac{R _{\text {out }}}{R _{\text {in }}}\right) $
where, $\beta$ is current gain.
$ \begin{array}{ll} \Rightarrow & \beta^{2}=A _P \times \frac{R _{\text {in }}}{R _{\text {out }}}=10^{6} \times \frac{100}{10 \times 10^{3}} \\ \Rightarrow & \beta^{2}=10^{4} \text { or } \beta=10^{2} \end{array} $
BETA
