Q 12-14-189JEE MainJEE Main 2019 (12 Jan, Shift 2)Medium
In the figure, given that $V_{BB}$ supply can vary from $0$ to $5.0$ V, $V_{CC} = 5$ V, $\beta_{dc} = 200$, $R_B = 100$ kΩ, $R_C = 1$ kΩ and $V_{BE} = 1.0$ V. The minimum base current and the input voltage at which the transistor will go to saturation, will be, respectively:
Answer: (C) $25\ \mu$A and $3.5$ V
At saturation $V_{CE} \approx 0$, so
$$I_C = \frac{V_{CC}}{R_C} = \frac{5}{1000} = 5\ \text{mA},\qquad I_B = \frac{I_C}{\beta} = \frac{5\ \text{mA}}{200} = 25\ \mu\text{A}$$
Input voltage:
$$V_{BB} = I_BR_B + V_{BE} = 25\times10^{-6}\times10^5 + 1 = 3.5\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics