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(A) In a transistor,the base region is designed to be very thin and lightly doped. This configuration minimizes the recombination of charge carriers (

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to enable the collector to collect $95\%$ of the holes or electrons coming from the emitter side

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(A) In a transistor,the base region is designed to be very thin and lightly doped. This configuration minimizes the recombination of charge carriers (holes or electrons) within the base region. As a result,the majority of charge carriers injected from the emitter can pass through the base and reach the collector. Specifically,this design enables the collector to collect approximately $95\%$ of the charge carriers coming from the emitter side.

In a transistor,the base is made very thin and is lightly doped with an impurity because:

Similar Questions

In the shown common emitter amplifier circuit,$\beta = 80$ and $V_{BE} = 0.7\,V$. The value of $R_C$ is:

Assertion: $A$ transistor amplifier in common emitter configuration has a low input impedance.
Reason: The base-to-emitter region is forward-biased.

The current gain of a transistor is $0.98$. If the transistor is used in a common emitter arrangement, what would be the change in collector current corresponding to a change of $0.5 \,mA$ in the base current (in $\,mA$)?

The input resistance of a common-emitter amplifier is $2 \, k\Omega$ and a.c. current gain is $20$. If the load resistor used is $5 \, k\Omega$,calculate the transconductance of the transistor used in $\Omega^{-1}$.

$A$ common emitter amplifier circuit is shown in the figure. For the transistor used in the circuit,the current amplification factor is $\beta_{dc} = 100$. Other parameters are given in the figure.

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