Question

The relationship between current gain \[\alpha \] in Common Base [CB] mode and current gain \[\beta \] in Common Emitter [CE] mode is
A. \[\beta = \alpha + 1\]
B. \[\beta = \dfrac{\alpha }{{1 - \alpha }}\]
C. \[\beta = \dfrac{\alpha }{{1 + \alpha }}\]
D. \[\beta = 1 - \alpha \]

Answer 1

Hint:Use the formulae for the current gain in Common Base mode and current gain in Common Emitter mode. Also use the relation between the emitter current, collector current and base current. Substitute the values of the current gains in Common Base mode and Common Emitter mode in this relation and solve it.

Formulae used:
The current gain \[\alpha \] in Common Base [CB] mode is given by
\[\alpha = \dfrac{{{I_c}}}{{{I_e}}}\] …… (1)
Here, \[{I_c}\] is the current from the collector and \[{I_e}\] is the current form emitter.
The current gain \[\beta \] in Common Emitter [CE] mode is given by
\[\beta = \dfrac{{{I_c}}}{{{I_b}}}\] …… (2)
Here, \[{I_c}\] is the current from the collector and is the current form base.\[{I_b}\]

Complete step by step answer:
We have asked to determine the relationship between the current gain \[\alpha \] in Common Base [CB] mode and the current gain \[\beta \] in Common Emitter [CE] mode.From equation (1). We can write
\[\dfrac{1}{\alpha } = \dfrac{{{I_e}}}{{{I_c}}}\]
From equation (2). We can write
\[\dfrac{1}{\beta } = \dfrac{{{I_b}}}{{{I_c}}}\]
We know that in a transistor the sum of the current from the collector and the current from base is equal to the current from the emitter.
\[{I_e} = {I_c} + {I_b}\]
Multiply both sides of the above equation by the collector current \[{I_c}\].
\[\dfrac{{{I_e}}}{{{I_c}}} = 1 + \dfrac{{{I_b}}}{{{I_c}}}\]

Substitute \[\dfrac{1}{\alpha }\] for \[\dfrac{{{I_e}}}{{{I_c}}}\] and \[\dfrac{1}{\beta }\] for \[\dfrac{{{I_b}}}{{{I_c}}}\] in the above equation.
\[\dfrac{1}{\alpha } = 1 + \dfrac{1}{\beta }\]
\[ \Rightarrow \dfrac{1}{\beta } = \dfrac{1}{\alpha } - 1\]
\[ \Rightarrow \dfrac{1}{\beta } = \dfrac{{1 - \alpha }}{\alpha }\]
\[ \therefore \beta = \dfrac{\alpha }{{1 - \alpha }}\]
Therefore, the relationship between the current gain \[\alpha \] in Common Base [CB] mode and the current gain \[\beta \] in Common Emitter [CE] mode is \[\beta = \dfrac{\alpha }{{1 - \alpha }}\].

Hence, the correct option is B.

Note: The students should be careful while using the formulae for the current gain in Common Base mode and Common Emitter mode. If the ratio of the currents in these formulae is taken incorrectly then we will also end with the incorrect relation between the current gain in Common Base and Common Emitter mode.