Question

An NPN transistor is used in common emitter configuration as an amplifier with $1k\Omega$ load resistance. Signal voltage of $10mV$ is applied across the base-emitter. This produces a $3mA$ 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. $0.33k\Omega ,1.5$
• B. $0.67k\Omega ,200$
• C. $0.33k\Omega ,300$
• D. $0.67k\Omega ,300$

Answer 1

Answer: (D)

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