Question

A transistor is connected in a common emitter (CE) configuration. The collector supply is $8V$ and the voltage drop across a resistor of $800\Omega $ in the collector circuit is $0.8V$. If the current gain factor $\left( \alpha \right)$ is$0.96$, then the change in the base current is
 (A) $\dfrac{1}{{24}}mA$
(B) $\dfrac{1}{{12}}mA$
(C) $\dfrac{1}{9}mA$
(D) $\dfrac{1}{3}mA$

Answer 1

Hint: The configuration within which the electrode is connected between the collector and base is known as a common emitter configuration. In this the input circuit gets connected to the emitter and also the output circuit also gets connected to this. Thus, the emitter is common to each input and therefore the output circuit, and thus the name is that of the common emitter configuration.

Formula used:
Collector current,
$ \Rightarrow {I_c} = \dfrac{{{V_{CE}}}}{{{R_E}}}$
Where, ${I_C}$ is the collector current, ${V_{CE}}$ is the voltage through CE configuration, and ${R_E}$ is the resistance.
Current gain,
$ \Rightarrow \beta = \dfrac{{{I_C}}}{{{I_B}}}$
Where, $\beta $ is the current gain, ${I_C}$ and ${I_B}$ is the base and collector current respectively.

Complete step by step Solution: This question is based on the CE configuration so before solving this we should know of this. If not then we will make you understand through this question. It is given that it is CE configuration and the voltages and resistance are given. We have to find the base current. So for this, we will first calculate the collector current.
So the collector current will be,
$ \Rightarrow {I_c} = \dfrac{{{V_{CE}}}}{{{R_E}}}$
Putting the values, we will get
$ \Rightarrow \dfrac{{8V}}{{800\Omega }}$
$ \Rightarrow 1mA$
Since the current gain is equal to
$ \Rightarrow \beta = \dfrac{{{I_C}}}{{{I_B}}}$
And also further expanding the formula we will get, the current gain which will be equal to
$ \Rightarrow \beta = \dfrac{\alpha }{{1 - \alpha }}$
Putting the values from the question and we get,
$ \Rightarrow \dfrac{{0.96}}{{1 - 0.96}}$
$ \Rightarrow 24$
Therefore now we will be able to calculate the ${I_B}$
Since,
$ \Rightarrow {I_B} = \dfrac{{{I_C}}}{\beta }$
Putting the values we will get,
$ \Rightarrow \dfrac{1}{{24}}mA$

Note: The characteristic of the common emitter transistor circuit is shown within the figure below. The emitter voltage is varied by using the potentiometer. And therefore the collector to emitter voltage varied by adjusting the potentiometer ${R_2}$. For the varied setting, this and voltage are taken from the millimeters and meter.