Question

A npn transistor in a common emitter mode is used as a simple voltage amplifier with a collector current of $4 mA$. the terminal of a $8 V$ battery is connected to the collector through a load resistance $R_L$ and to the base through a resistance $R_{ B }$. The collector emitter voltage $V_{C E}=4 V$, baseemitter voltage $V_{ BE }=0.6$ and the base current amplification factor $\beta_{ dc }=100$, calculate the value of $R_B$.

• A. $R _{ B }=15\, k \Omega$
• B. $R _{ B }=200\, k \Omega$
• C. $R _{ B }=1\, k \Omega$
• D. $R _{ B }=185\, k \Omega$

Answer 1

Answer: (D)

By the definition

$\therefore i_B=\frac{i_C}{\beta_{d c}}=\frac{4 m A}{100}=0.04\, mA$
From figure
$i_B R_B+V_{B E}=V_{C C} $
$\therefore R_B=\frac{V_{C C}-V_{B E}}{i_B}=\frac{(8-0.6)}{0.04 \times 10^{-3}}=185 \,k \Omega $