Question

A transistor is connected in common-emitter $\left(\text{C.E}\right)$ configuration. The collector supply is $8\text{V}$ and the voltage drop across a resistor of $800\Omega$ in the collector circuit is $0.5\text{V}.$. If the current gain factor $\left(\alpha \right)$ is $\mathrm{0.96.}$ The base-current is $(20+n)\mu \text{A}$. The value of $n$ is.(integer only)

• A. 6.0
• B. 6
• C. 6.00

Answer 1

Answer: (A), (B), (C)

Given, $\alpha =0.96;V_{CE}=8\text{V}{R}_L=800\Omega ;V_L=0.5\text{V}$

The current amplification in $\left(\text{C.E}\right)$ mode is,

$\beta =\frac{\alpha }{1-\alpha }=\frac{0.96}{1-0.96}=24$

The collector-current is,

$i_c=\frac{\text{Voltage-drop across resistor}\left(V_L\right)}{\text{resistance}\left(R_L\right)}$

$=\frac{0.5}{800}=0.625\times {10}^{-3}\text{A}$

$\because \beta =\frac{i_c}{i_B}$

Where, $i_B$ is base current

$\therefore i_B=\frac{i_c}{\beta }=\frac{0.625\times {10}^{-3}\text{A}}{24}=26\times {10}^{-6}\text{A}=26\mu \text{A}$

$\therefore i_B=(20+6)\mu \text{A}$

$\therefore n=6$