Question

An NPN 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 $15\frac{1}{4}\,A$ change in the base current of the amplifier. The input resistance and voltage gain are :

• A. $0.33 \,k\Omega,\, 1.5$
• B. $0.67\, k\,\Omega,\, 200$
• C. $0.33 k\,\Omega,\, 300$
• D. $0.67 k\,\Omega,\, 300$

Answer 1

Answer: (D)

input current $= 15 × 10^{-6}$
output current $= 3 × 10^{-3}$
resistance output = 1000
$V_{input} = 10 \times 10^{-3}$
Now $V_{input} = r_{input} \times i_{input}$
$10 \times 10^{-3} = r_{input} \times 15 \times 10^{-6}$
$r_{input} \, = \frac{2000}{3} \, = \, 0.67 K \Omega$
voltage gain = $\frac{V_{output}}{V_{input}} = \frac{1000 \times 3 \times 10^{-3}}{10 \times 10^{-3}} = 300$