Question

In the common base circuit of a transistor, current amplification factor is $0.95$. Calculate the emitter current if base current is $0.2mA$.
A) 2mA
B) 4mA
C) 6mA
D) 8mA

Answer 1

Transistor: It is an electronic component used in a circuit to control a large amount of current or voltage with a small amount of voltage or current.
It is a semiconductor device that can be used to amplify or switch (rectify) electrical signals or power, allowing it to be used in a wide array of electronic devices.
If a voltage or current is applied to one pair of the transistors terminals that control the current will flow through another pair of terminals in the configuration.
The value of controlled output power is higher than the controlling input power; a transistor can amplify a signal.
Some transistors are individually packed, but some transistors are found embedded in integrated circuits.

Formula used:
$\alpha = \dfrac{{{\text{collector current}}}}{{{\text{emitter current}}}}$,
$\alpha = \dfrac{{{{\text{I}}_{\text{c}}}}}{{{{\text{I}}_{\text{e}}}}}$

Complete step by step solution:
It is given that the data are $\alpha = 0.95$, and base current, ${{\text{I}}_{\text{b}}} = 0.2{\text{mA}}$
Then we have to find out the emitter current.
So we have to use the formula and we write it as, $\alpha = \dfrac{{{{\text{I}}_{\text{e}}} - {{\text{I}}_{\text{b}}}}}{{{{\text{I}}_{\text{e}}}}}$
Here we have to rewrite the formula we get,
$\Rightarrow \alpha = 1 - \dfrac{{{{\text{I}}_{\text{b}}}}}{{{{\text{I}}_{\text{e}}}}}$
First we have to find $\alpha$;
So we use the formula and we get,
$\alpha = 1 - \alpha$
Putting the value of $\alpha$ and we get
$\Rightarrow 1 - 0.95$
Let us subtract it as,
$= 0.05$
Now, we can determine the value of ${{\text{I}}_{\text{e}}}$,
${{\text{I}}_{\text{e}}} = \dfrac{{{{\text{I}}_b}}}{\alpha }$
Putting the values and we get,
${{\text{I}}_{\text{e}}} = \dfrac{{0.2}}{{0.05}}$
Let us divide the term we get = $4{\text{mA}}$

Hence the correct option is (B).

Note: In the common-base configuration, the collector current $\;\left( {{I_c}} \right)$ acts as the output current, and the emitter current $\left( {{I_e}} \right)$ acts as the input current.
The current amplification factor is defined as the ratio of change in emitter current to the collector current at a constant collector-base voltage of a transistor in the common base configuration.
In the numerical analysis, an amplification factor is a number derived by using Von Neumann stability analysis to determine the stability of a numerical scheme for a partial differential equation.
The common emitter amplifiers are used to determine the large current gain.
The input signal is applied between the base and emitter terminals of the transistor and the output signal is obtained between the collector and emitter terminals of the configuration.