Q 12-14-183JEE MainJEE Main 2019 (8 Apr, Shift 2)Easy
A common emitter amplifier circuit, built using an NPN transistor, has its base connected through $R_B$ to a supply $V_B$ and its collector connected through $R_C$ to $V_{CC}$. Its dc current gain is $250$, $R_C = 1\ \text{k}\Omega$ and $V_{CC} = 10\ \text{V}$. The minimum base current for $V_{CE}$ to reach saturation is
Answer: (B) $40\ \mu\text{A}$
At saturation $V_{CE} \approx 0$, so the collector current is limited by $R_C$:
$$I_C = \frac{V_{CC}}{R_C} = \frac{10}{1000} = 10\ \text{mA}$$
$$I_B = \frac{I_C}{\beta} = \frac{10\ \text{mA}}{250} = 40\ \mu\text{A}$$
Solution by Sreeraj P, M.Sc Physics