Question

An $NPN$ transistor is used in common emitter configuration as an amplifier with $1\text{k}\Omega$ load resistance. Signal voltage of $10\text{mV}$ is applied across the base-emitter. This produces a $3\text{mA}$ change in the collector current and $15\mu \text{A}$ change in the base current of the amplifier. The input resistance and voltage gain are

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

Answer 1

The correct option is D $0.67\text{k}\Omega ,300$

Given:
$\Delta V_i=10\times {10}^{-3}\text{V}$
$\Delta I_c=3\times {10}^{-3}\text{A}$
$\Delta I_b=15\times {10}^{-6}\text{A}$
Resistence
$R_i=\frac{\Delta V_i}{\Delta I_b}$

$R=\frac{10\times {10}^{-3}}{15\times {10}^{-6}}=0.67\text{k}\Omega$

$\therefore$ Voltage gain $=\frac{\Delta I_c}{\Delta I_b}\times \frac{R_0}{R_i}=\left(\frac{3\times {10}^{-3}}{15\times {10}^{-6}}\right)\times \frac{1000}{670}=200\times \frac{1000}{670}\approx 300$

Hence option $\left(B\right)$ is correct.