Modern Physics 6 Question 7
7. An $n-p-n$ transistor is used in common emitter configuration as an amplifier with $1 k \Omega$ load resistance. Signal voltage of $10 mV$ is applied across the base-emitter. This produces a 3 $mA$ change in the collector current and $15 \mu A$ change in the base current of the amplifier. The input resistance and voltage gain are
(a) $0.67 k \Omega, 200$
(b) $0.33 k \Omega, 1.5$
(c) $0.67 k \Omega, 300$
(d) $0.33 k \Omega, 300$
(Main 2019, 9 April I)
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Solution:
- Given, load resistance, $R _L=1 k \Omega$
Input voltage, $V _{\text {in }}=10 mV=10 \times 10^{-3} V$
Base current, $\Delta I _B=15 \mu A=15 \times 10^{-6} A$
Collector current, $\Delta I _C=3 mA$
Input resistance,
$$ R _{in}=\frac{V _{in}}{\Delta I _B}=\frac{10 \times 10^{-3}}{15 \times 10^{-6}}=0.67 k \Omega $$
and voltage gain $=\beta \times \frac{R _L}{R _{\text {in }}}=\frac{\Delta I _C \times R _L}{\Delta I _B \times R _{\text {in }}}$
$$ \begin{aligned} & =\frac{3 mA}{15 \mu A} \times \frac{1 k \Omega}{0.67 k \Omega} \\ & =\frac{3 \times 10^{-3}}{15 \times 10^{-6}} \frac{1 \times 10^{3}}{0.67 \times 10^{3}} \\ & =\frac{1000 \times 3 \times 3}{15 \times 2}=300 \quad(\because 0.67 \cong 2 / 3) \end{aligned} $$
Alternate Solution
$\therefore$ Voltage gain $=\frac{V _{\text {output }}}{V _{\text {input }}}=\frac{R _L \times \Delta U _C}{V _{\text {in }}}$
$$ =\frac{1 \times 10^{3} \times 3 \times 10^{-3}}{10 \times 10^{-3}}=300 $$
