An npn transistor operates as a common emitte | vedclass

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(B) The power gain $(A_p)$ of a common emitter amplifier is given by the formula: $A_p = \beta^2 	imes \frac{R_o}{R_i}$, where $\beta$ is the current

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(B) The power gain $(A_p)$ of a common emitter amplifier is given by the formula: $A_p = \beta^2 	imes \frac{R_o}{R_i}$, where $\beta$ is the current gain, $R_o$ is the output resistance, and $R_i$ is the input resistance.Given: $A_p = 10^6$, $R_i = 100\, \Omega$, and $R_o = 10\, k\Omega = 10^4\, \Omega$.Substituting the values into the formula:$10^6 = \beta^2 	imes \frac{10^4}{10^2}$$10^6 = \beta^2 	imes 10^2$$\beta^2 = \frac{10^6}{10^2} = 10^4$$\beta = \sqrt{10^4} = 100$.Thus, the common emitter current gain $\beta$ is $100$.

An $npn$ transistor operates as a common emitter amplifier with a power gain of $10^{6}$. The input circuit resistance is $100\, \Omega$ and the output load resistance is $10\, k\Omega$. The common emitter current gain $\beta$ will be (Round off to the nearest integer).

Similar Questions

For a given transistor amplifier circuit in $CE$ configuration with $V_{CC} = 1 \text{ V}$,$R_c = 1 \text{ k}\Omega$,$R_b = 100 \text{ k}\Omega$,and $\beta = 100$,the value of the base current $I_b$ is:

In the $CB$ mode of a transistor, when the collector voltage changes by $0.5 \ V$, the collector current changes by $0.05 \ mA$. What is the output resistance in $k\Omega$?

$Assertion$ : In common base configuration, the current gain of the transistor is less than unity.
$Reason$ : The collector terminal is reverse biased for amplification.

In which of the configurations of a transistor, the power gain is highest?

Which of the following regions of transistors are,respectively,heavily doped and lightly doped?

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