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An npn transistor operates as a common emmitter amplifier, with a power gain of

60 dB. The input circuit resistance is $100\Omega$ and the output load resistance is $10k\Omega$ .

The common emitter cuttent gain $\beta$ is :

Option 1)
$10^{2}$

Option 2)
$60$

Option 3)
$6\times 10^{2}$

Option 4)
$10^{4}$

Answers (1)

best_answer

Relation between α and β -

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

- wherein

$\alpha = \frac{I_{C}}{I_{E}}$

$\beta = \frac{I_{C}}{I_{B}}$ (current gain )

$60dB=10\log_{10}(\frac{P}{P_o})$

$(\frac{P}{P_o})=10^{6}=power\: \: gain$

$power\: \: gain=(\beta )^{2}\times \frac{R_{out}}{R_{in}}$

$\beta = current\: \: gain$

So,

$(\beta )^{2}=10^{6}\times \frac{R_{in}}{R_{out}}=10^{6}\times \frac{10^{2}}{10^{4}}$

$\beta=100\: \: or\: \: 10^{2}$

Option 1)

$10^{2}$

Option 2)

$60$

Option 3)

$6\times 10^{2}$

Option 4)

$10^{4}$

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