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Question: The current gain in a common base circuit is 40. The ratio of emitter current to base current is ...

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

Explanation

Solution

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, β=40\beta = 40.
We will now express the current amplification factor in terms of current gain. So, we can say that,
α=β1+β\alpha = \dfrac{\beta }{{1 + \beta }} ………...… (I)
Here, α\alpha is the current amplification factor, and β\beta is the current gain factor.
We can now substitute the value of β=40\beta = 40in equation (I) to find the value of α\alpha . So we get,
α=401+40\Rightarrow \alpha = \dfrac{{40}}{{1 + 40}}
α=4041\Rightarrow \alpha = \dfrac{{40}}{{41}}
We will now write the expression for α\alpha and β\beta in terms of collector current, base current and emitter current as,
β=IcIb\beta = \dfrac{{{I_c}}}{{{I_b}}}
Here, Ic{I_c} and Ib{I_b} are the collector and base currently respectively.
We will solve it further and it gives,
Ic=βIb\Rightarrow {I_c} = \beta {I_b} …………... (II)
Similarly, we can say that,
α=IcIe\alpha = \dfrac{{{I_c}}}{{{I_e}}}
Here, Ie{I_e} is the emitter current.
We will solve it further and it will give,
Ic=Ieα\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,
Ieα=Ibβ{I_e}\alpha = {I_b}\beta
IeIb=βα\Rightarrow \dfrac{{{I_e}}}{{{I_b}}} = \dfrac{\beta }{\alpha } ………………... (IV)
We can now substitute.α=4041\alpha = \dfrac{{40}}{{41}} and β=40\beta = 40 in equation (IV) to find the required value of IeIb\dfrac{{{I_e}}}{{{I_b}}}
IeIb=404041=  40×4140\Rightarrow \dfrac{{{I_e}}}{{{I_b}}} = \dfrac{{40}}{{\dfrac{{40}}{{41}}}}\, = \;\dfrac{{40 \times 41}}{{40}}
IeIb=41\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 1mA1\,{\rm{mA}}, it is about 26  Ω26\;\Omega and for 2mA2\,{\rm{mA}} it is about 13  Ω13\;\Omega .