Question

In a common emitter transistor amplifier, signal voltage across the collector resistance of $2k\Omega $ is $2volt$ . What should be the value of resistance $R_{B}$ in series with $V_{B B}$ voltage supply of $2volt$ in the input circuit, if the $D.C.$ base current is $10$ times the signal current in $R_{B}$ ?
$\left(\right.$ Given: base emitter voltage $\text{V}_{B E}=0.6V$ , current amplification factor $\beta _{a c} \approx \beta _{d c} = 100)$

• A. $7k\Omega $
• B. $26k\Omega $
• C. $10k\Omega $
• D. $14k\Omega $

Answer 1

Answer: (D)

$\left( i _{ C }\right)_{ AC }=\frac{\left( V _{\text {out }}\right)_{ AC }}{ R _{ C }}=\frac{2 volt }{2 k \Omega}=1 mA$
$\left( i _{ B }\right)_{ AC }=\frac{\left( i _{ C }\right)_{ AC }}{\beta_{ ac }}=\frac{1 mA }{100}=0.01 mA$
$\left( i _{ B }\right)_{ DC }=10\left( i _{ B }\right)_{ AC }=0.1 mA =10^{-4} A$
$R_{B}=\frac{V}{i_{B}}=\frac{2-0.6}{10^{-4}}=14 k \Omega$