Q124 / 125 · BITSAT 2014 · Physics
For a CE transistor amplifier, the audio signal voltage across the collector resistance of $2.0 \mathrm{k} \Omega$ is $2.0 \mathrm{~V}$. Suppose the current amplification factor of the transistor is 100 , What should be the value of $\mathrm{R}_{\mathrm{B}}$ in series with $\mathrm{V}_{\mathrm{BB}}$ supply of $2.0 \mathrm{~V}$ if the dc base current has to be 10 times the signal current?
- Correct answerA$14 \mathrm{k} \Omega$
- B$18 \mathrm{k} \Omega$
- C$10 \mathrm{k} \Omega$
- D$5 \mathrm{k} \Omega$
Solution
The output ac voltage is $2.0 \mathrm{~V}$. So, the ac collector current $\mathrm{i}_{\mathrm{C}}=2.0 / 2000=1.0 \mathrm{~mA}$.
The signal current through the base is, therefore given by $\mathrm{i}_{\mathrm{B}}=\mathrm{i}_{\mathrm{C}} / \beta=1.0 \mathrm{~mA} / 100=0.010 \mathrm{~mA}$
The dc base current has to be $10 \times 0.010$ $=0.10 \mathrm{~mA}$
$$
\mathrm{R}_{\mathrm{B}}=\left(\mathrm{V}_{\mathrm{BB}}-\mathrm{V}_{\mathrm{BE}}\right) / \mathrm{I}_{\mathrm{B}}
$$
Assuming $\mathrm{V}_{\mathrm{BE}}=0.6 \mathrm{~V}, \mathrm{R}_{\mathrm{B}}=(2.0 .0 .6) / 0.10$
$=14 \mathrm{k} \Omega$
Hence, the correct answer is option: 'A'.