Question

The transfer characteristic curve of a transistor, having input and output resistance \(100~\Omega\) and \(100~k\Omega\) respectively, is shown in the figure, The voltage and power gain, are respectively
L, (na) —

• A. $5\times {10}^4,5\times {10}^6$
• B. $5\times {10}^4,2.5\times {10}^6$
• C. $2.5\times {10}^4,2.5\times {10}^6$
• D. $2\times {10}^4,4\times {10}^5$

Answer 1

The correct option is D $2\times {10}^4,4\times {10}^5$

We know that,
$\beta =\frac{\Delta i_c}{\Delta i_b}$

From the graph,
$\beta =\frac{200-100}{10-5}=20$

$\text{voltage gain}{A}_V=\beta \frac{R_o}{R_i}$

$\Rightarrow A_V=\frac{20\times 100\times {10}^3}{100}$

$\Rightarrow A_V=2\times {10}^4$

$\text{Power gain}{A}_P={\beta }^2\frac{R_o}{R_i}$

$\Rightarrow A_P={20}^2\left(\frac{100\times {10}^3}{100}\right)$

$\Rightarrow A_P=4\times {10}^5$

Hence, option $\left(C\right)$ is correct.