Question

An $\text{n-p-n}$ transistor in a common emitter mode is used as a simple voltage amplifier as shown in figure. Calculate the value of $R_B$. [Given: $\beta =100$]

• A. $1.85\times {10}^5\Omega$
• B. $1.85\times {10}^3\Omega$
• C. $1.85\times {10}^2\Omega$
• D. $1.85\times {10}^4\Omega$

Answer 1

The correct option is A $1.85\times {10}^5\Omega$

We know, $\beta =\frac{I_C}{I_B}$

$\Rightarrow 100=\frac{4\times {10}^{-3}}{I_B}$

$\Rightarrow I_B=4\times {10}^{-5}\text{A}$

Now, potential difference across $R_B$ is

$V_{R_B}=I_B{R}_B=8-V_{BE}=8-0.6=7.4\text{V}$

$\Rightarrow R_B=\frac{7.4}{I_B}$

$=\frac{7.4}{4\times {10}^{-5}}=1.85\times {10}^5\Omega$

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