
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
571.5k+ views
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 $.
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 $.
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Differentiate between action potential and resting class 12 biology CBSE

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Explain sex determination in humans with the help of class 12 biology CBSE

