Question

For a CE transistor amplifier, the audio signal voltage across the collector resistance of $2k\Omega$ is $2V$. Suppose the current amplification factor of the transistor is $100$. What should be the value of $R_B$ in series with $V_{BB}$ supply of $2V$ if the dc base current has to be $10$ times the signal current. Also calculate the dc drop across the collector resistance. Take $V_{BE}=0.6V$

Answer 1

$RB=14k\Omega$
$dc=20$ volt
Current at collector due to audio $s/g=\frac{2v}{2k\Omega }=1mA$
$I_C=1mA$
If base current $\left(I_B\right)$ is $10$ times, signal current mean,
$I_B=10\mu Amp$
$\beta =100$, signal current at base input will be $=\frac{1mA}{100}=\frac{I_b}{\beta }=10\mu Amp$
By considering the $V_{BE}=0.6$ Volt, $R_B$ value for $100\mu A$ is given by
$R_B=\frac{1.4}{100\times {10}^{-6}}=14k\Omega$