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)

 

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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