In the figure, given that $V_{BB}$ supply can vary from 0 to 5.0 V, $V_{cc}=5 V$, $\beta_{dc}=200$, $R_B=100 k\Omega$, $R_C=1 k\Omega$ 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:

Semiconductors — JEE Main · 2019 · 12 Jan 2019 · Shift 2 Question with Solution

In the figure, given that

V_{B B}

supply can vary from 0 to 5.0 V,

V_{cc} = 5 V

β_{d c} = 200

R_{B} = 100 k Ω

R_{C} = 1 k Ω

and

V_{B E} = 1.0 V

. The minimum base current and the input voltage at which the transistor will go to saturation, will be respectively:

Answer: (A)

At saturation the collector

-

emitter voltage collapses to zero, so the full supply appears across

R_{C}

V_{C C} = I_{C} R_{C}

, giving

I_{C} = \frac{5}{1000} = 5 \times 1 0^{- 3}

A. The base current needed to drive this is

i_{B} = \frac{I _{C}}{β} = \frac{5 \times 1 0 ^{- 3}}{200} = 25 μ

A. Taking Kirchhoff's voltage law round the input loop,

V_{B B} = i_{B} R_{B} + V_{B E} = (25 \times 1 0^{- 6}) (100 \times 1 0^{3}) + 1.0 = 2.5 + 1.0 = 3.5

V, so the transistor just reaches saturation with a base current of

25 μ

A at an input of

3.5

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option