Question

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

• A. $20\,\mu A$ and $3.5\,V$
• B. $25\,\mu A$ and $3.5\,V$
• C. $25\,\mu A $ and $2.5\,V$
• D. $20\,\mu A$ and $2.8\,V$

Answer 1

The Correct Option is B

At saturation, V$_{CE}$ = 0
$V_{CE} \, = \, V_{CC} \, - \, I_C R_C$
$\Rightarrow \, \, I_C \, = \frac{V_{CC}}{R_C} \, = \, 5 \, \times \, 10^{-3} \, A$
Given
$\beta_{dc} \, = \, \frac{I_C}{I_B}$
$I_B \, = \, \frac{5 \times \, 10^{-3}}{200}$
$I_B \, = \, 25 \, \mu A$
At input side
$V_{BB} \, = \, I_B \, R_B \, + \, V_{BE}$
$ \, \, \, \, =(25 mA)(100 k \Omega) \, + \, 1V$
$V_{BB} \, \, = \, \, 3.5 \, V$