In the figure, given that VBB supply can vary from 0 to 5.0 V, VCC =… | JEE Main 2019 (12 Jan, Shift 2)

Q88JEE Main 2019 · 12 JAN · SHIFT 2Difficult

Junction Transistor

In the figure, given that VBB supply can vary from 0 to 5.0 V, VCC = 5V, $\beta$ dc = 200, RB = 100 k $\Omega$ , RC = 1 k $\Omega$ and VBE = 1.0 V. The minimum base current and the input voltage at which the transistor will go to saturation, will be respectively:

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Answer: (d)

At saturation $V_{CE} \approx 0$ : $I_C = \dfrac{V_{CC}}{R_C} = \dfrac{5\ \text{V}}{1\ \text{k}\Omega} = 5\ \text{mA}$
Minimum base current: $I_B = \dfrac{I_C}{\beta} = \dfrac{5\ \text{mA}}{200} = 25\ \mu\text{A}$
Input voltage: $V_{BB} = I_B R_B + V_{BE} = 25\ \mu\text{A} \times 100\ \text{k}\Omega + 1 = 3.5\ \text{V}$
So the answer is (d) $25\ \mu\text{A}$ and 3.5 V.

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