Question

A common emitter amplifier circuit, built using an $\text{npn}$ transistor, is shown in the figure. Its dc current gain is $250,R_C=1\text{k}\Omega$ and $V_{CC}=10\text{V}$. What is the minimum base current for $V_{\text{CE}}$ to reach saturation ?

• A. $40\mu \text{A}$
• B. $100\mu \text{A}$
• C. $10\mu \text{A}$
• D. $7\mu \text{A}$

Answer 1

The correct option is A $40\mu \text{A}$

In the saturation state, $V_{CE}$ becomes equal to zero.

$\Rightarrow I_C=\frac{V_{CC}}{R_C}=\frac{10}{1000}=10\text{mA}$

Now, $\beta =\frac{I_C}{I_B}\Rightarrow I_B=\frac{I_C}{\beta }$

$\therefore I_B=\frac{10\times {10}^{-3}}{250}=40\mu \text{A}$

Hence, $\left(A\right)$ is the correct answer.