Modern Physics Ques 227
- The transfer characteristic curve of a transistor, having input and output resistance $100 $ $\Omega$ and $100 $ $k \Omega$ respectively, is shown in the figure. The voltage and power gain, are respectively,
(Main 2019, 12 April I)
(a) $2.5 \times 10^{4}, 2.5 \times 10^{6}$
(b) $5 \times 10^{4}, 5 \times 10^{6}$
(c) $5 \times 10^{4}, 5 \times 10^{5}$
(d) $5 \times 10^{4}, 2.5 \times 10^{6}$
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Answer:
Correct Answer: 227.(d)
Solution:
Formula:
- Given curve is between $I _c$ and $I _b$ as output and input currents, respectively
So, it is transfer characteristics curve of a common emitter (CE) configuration.
In $CE$ configuration,
Current gain, $\beta=\frac{I _{\text {out }}}{I _{\text {in }}}=\frac{I _c}{I _b} \quad …….(i)$
Voltage gain,
$ A _V=\frac{V _{\text {out }}}{V _{\text {in }}}=\frac{I _c \times R _{\text {out }}}{I _b \times R _{\text {in }}}=\beta \times \frac{R _{\text {out }}}{R _{\text {in }}} \quad …….(ii) $
and power gain
$ A _P=\frac{P _{\text {out }}}{P _{\text {in }}}=\frac{I _c^{2} \times R _{\text {out }}}{I _b^{2} \times R _{\text {in }}}=\beta^{2} \times \frac{R _{\text {out }}}{R _{\text {in }}} \quad …….(iii) $
Given, $R _{\text {in }}=100 $ $\Omega$ and $R _{\text {out }}=100 \times 10^{3} $ $\Omega$
From Eq. (i), we get
$ \begin{aligned} \beta & =\frac{5 mA}{100 \mu A}\left( \text { or } \frac{10 mA}{200 \mu A} \text { or } \frac{15 mA}{300 \mu A} \text { or } \frac{20 mA}{400 \mu A}\right) \\ \Rightarrow \quad \beta & =\frac{5 \times 10^{-3}}{100 \times 10^{-6}}=50 \quad …….(iv) \end{aligned} $
From Eqs. (ii) and (iv), we get
Voltage gain, $A _V=\beta \times \frac{R _{\text {out }}}{R _{\text {in }}}=50 \times \frac{100 \times 10^{3}}{100}$
$\Rightarrow \quad $ $ A _V=50000=5 \times 10^{4} $
From Eqs. (iii) and (iv), we get
Power gain, $A _P=\beta^{2} \times \frac{R _{\text {out }}}{R _{\text {in }}}$
$ \begin{aligned} & =(50)^{2} \times \frac{100 \times 10^{3}}{100} \\ & =2500 \times 1000 \\ \Rightarrow \quad A _P & =2.5 \times 10^{6} \end{aligned} $
BETA
