Question

The current gain in a common base circuit is 40. The ratio of emitter current to base current is
A. 40
B. 41
C. 42
D. 43

Answer 1

Hint: As we all know that the common base amplifier is one of the types of Bipolar junction transistor where the terminal’s base is the common base for both the output and the input signals. It is used with the amplifier having fewer impedance levels.

Complete step by step answer:
It is given to us that the current gain, $\beta = 40$.
We will now express the current amplification factor in terms of current gain. So, we can say that,
$\alpha = \dfrac{\beta }{{1 + \beta }}$ ………...… (I)
Here, $\alpha $ is the current amplification factor, and $\beta $ is the current gain factor.
We can now substitute the value of $\beta = 40$in equation (I) to find the value of $\alpha $. So we get,
$\Rightarrow \alpha = \dfrac{{40}}{{1 + 40}}$
$\Rightarrow \alpha = \dfrac{{40}}{{41}}$
We will now write the expression for $\alpha $ and $\beta $ in terms of collector current, base current and emitter current as,
$\beta = \dfrac{{{I_c}}}{{{I_b}}}$
Here, ${I_c}$ and ${I_b}$ are the collector and base currently respectively.
We will solve it further and it gives,
$ \Rightarrow {I_c} = \beta {I_b}$ …………... (II)
Similarly, we can say that,
$\alpha = \dfrac{{{I_c}}}{{{I_e}}}$
Here, ${I_e}$ is the emitter current.
We will solve it further and it will give,
$ \Rightarrow {I_c} = {I_e}\alpha $ ……………. (III)
We will now equate equation (II) and equation (III) since both are the terms of collector current. Hence, we will get,
\[{I_e}\alpha = {I_b}\beta \]
\[ \Rightarrow \dfrac{{{I_e}}}{{{I_b}}} = \dfrac{\beta }{\alpha }\] ………………... (IV)
We can now substitute.$\alpha = \dfrac{{40}}{{41}}$ and $\beta = 40$ in equation (IV) to find the required value of \[\dfrac{{{I_e}}}{{{I_b}}}\]
\[\Rightarrow \dfrac{{{I_e}}}{{{I_b}}} = \dfrac{{40}}{{\dfrac{{40}}{{41}}}}\, = \;\dfrac{{40 \times 41}}{{40}}\]
$\therefore \,\,\dfrac{{{I_e}}}{{{I_b}}} = 41$

Therefore, we can see that the ratio of emitter current to base current is 41. Hence, the correct option is (B).

Note:
We must know that the input impedance of the common base circuit is too less for most source signals and it very strongly depends on collector current i.e. for $1\,{\rm{mA}}$, it is about $26\;\Omega $ and for $2\,{\rm{mA}}$ it is about $13\;\Omega $.