Question

In the circuit shown in figure, the current gain $\beta$ = 100 for a npn transistor. The bias resistance $R_B$ so that $V_{CE}$ = 5V is ($V_{BE}$ < < 10 V)

• A. $2\times {10}^3\Omega$
• B. ${10}^5\Omega$
• C. $2\times {10}^5\Omega$
• D. $5\times {10}^5\Omega$
• E. $1\times {10}^3\Omega$

Answer 1

The correct option is C $2\times {10}^5\Omega$

$\begin{array}{c}\beta =100\\ V_{CE}=5V\\ \beta =\frac{I_C}{I_B}=I_B=\frac{5_m}{100}\\ I_B=0.05mA\\ 10-I_{o}RB-V_{BE}=\mathrm{0......}\left(1\right)\\ 10-1K\times I_C-V_{CE}\mathrm{0........}\left(2\right)\\ I_C=\frac{5}{1K}=5mA\\ fromequation\\ if{V}_{BE}<<10\approx 0\approx V_{BE}\\ Hence,\\ R_B=\frac{10}{I_B}=2\times {10}^5\Omega \end{array}$

Therefore, option $C$ is correct.