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

Answer: (A)

$198\Omega$

$R_0=500k\Omega$; $\alpha =0.98$
Power gain $=6.0625\times {10}^6$
we have, voltage gain $A_V=\beta ,\frac{R_0}{R_i}$
But current gain,
$\beta =\frac{\alpha }{1-\alpha }=\frac{0.98}{1-0.98}=49$
$A_V=49\times \frac{500\times {10}^3}{R_i}=\frac{24.5\times {10}^6}{R_i}$
Given $6.0625\times {10}^6=A_V\times \beta$
$\left(\frac{24.5\times {10}^6}{6.0625}\right)$
or $R_i=\frac{24.5\times 49}{6.0625}=198\Omega$