An npn transistor operates as a common emitte | vedclass

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(B) The power gain in $dB$ is given by $G_p = 10 \log_{10} \left( \frac{P_{out}}{P_{in}} \right) = 60\, dB$.Therefore,$\frac{P_{out}}{P_{in}} = 10^{(6

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$10^2$

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(B) The power gain in $dB$ is given by $G_p = 10 \log_{10} \left( \frac{P_{out}}{P_{in}} ight) = 60\, dB$.Therefore,$\frac{P_{out}}{P_{in}} = 10^{(60/10)} = 10^6$.Power gain is also expressed as $A_p = \beta^2 	imes \frac{R_{out}}{R_{in}}$,where $\beta$ is the current gain,$R_{out} = 10\, k\Omega = 10^4\,\Omega$,and $R_{in} = 100\,\Omega$.Substituting the values: $10^6 = \beta^2 	imes \frac{10^4}{100}$.$10^6 = \beta^2 	imes 10^2$.$\beta^2 = \frac{10^6}{10^2} = 10^4$.$\beta = \sqrt{10^4} = 100$.

An $npn$ transistor operates as a common emitter amplifier,with a power gain of $60\, dB$. The input circuit resistance is $100\,\Omega$ and the output load resistance is $10\, k\Omega$. The common emitter current gain $\beta$ is:

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