Question

The input resistance of a common emitter transistor amplifier, if the output resistance is \[500k\Omega \], the current gain $\alpha = 0.98$ and power gain is $6.0625 \times {10^6}$ is:
A. $198\Omega $
B. $300\Omega $
C. $100\Omega $
D. $400\Omega $

Answer 1

Hint: We should remember the basic concepts and formula of a common emitter transistor amplifier i.e. $\delta V = \beta \dfrac{{{R_0}}}{{{R_i}}}$ and also the relation between the $\alpha $ and $\beta $ as $\beta = \dfrac{\alpha }{{1 - \alpha }}$.

Formula used – 1) $\Delta V = \beta \dfrac{{{R_0}}}{{{R_i}}}$
2) $\beta = \dfrac{\alpha }{{1 - \alpha }}$
3) $\Delta P = \beta \times \Delta V$

Complete Step-by-Step solution:
Given, ${R_0} = 500\Omega $
$\alpha = 0.98$ and $\Delta P = 6.0625 \times {10^6}$
Voltage gain, $\Delta V = \beta \dfrac{{{R_0}}}{{{R_i}}}$
Also, $\beta = \dfrac{\alpha }{{1 - \alpha }}$$ = \dfrac{{0.98}}{{1 - 0.98}}$=49
$\Delta V = (49)(\dfrac{{500}}{{{R_i}}})$
\[Power{\text{ }}gain\;(\Delta P) = Current\;gain(\beta ) \times voltage\;gain(\Delta V)\]
\[\
  6.0625 \times {10^6} = 49 \times 49 \times (\dfrac{{500}}{{{R_i}}}) \\
   \Rightarrow {R_i} = 198\Omega \\
\]
Hence, the correct answer is option A.

Note- In common emitter transistor amplifier, we should know that the $\beta = \dfrac{\alpha }{{1 - \alpha }}$ and the relationship between power gain, current gain and voltage gain i.e. \[Power{\text{ }}gain\;(\Delta P) = Current\;gain(\beta ) \times voltage\;gain(\Delta V)\].