Question

Obtain the relation between the current amplification factors $\alpha $ and $\beta $ of a transistor.

Answer 1

Hint: Here, we will proceed by writing down the various configurations possible of a transistor. Then, we will define the current amplification factors $\alpha $ and $\beta $. Finally, we will use the relation between emitter, collector and base current.

Formula used:
$\alpha = \dfrac{{\Delta {{\text{I}}_{\text{C}}}}}{{\Delta {{\text{I}}_{\text{E}}}}}$, $\beta = \dfrac{{\Delta {{\text{I}}_{\text{C}}}}}{{\Delta {{\text{I}}_{\text{B}}}}}$ and ${{\text{I}}_{\text{E}}} = {{\text{I}}_{\text{B}}} + {{\text{I}}_{\text{C}}}$.

Complete answer:
Any transistor consists of three terminals i.e., the emitter, the base and the collector. With the help of these three terminals, the transistor can be connected in a circuit with one of these terminals to be common to both the input and the output in three different possible configurations which are Common Base (CB) configuration, Common Emitter (CE) configuration, Common Collector (CC) configuration. In all the configurations, the emitter junction is forward biased and the collector junction is reverse biased.
The ratio of the change in the collector current ($\Delta {{\text{I}}_{\text{C}}}$) to the change in the emitter current ($\Delta {{\text{I}}_{\text{E}}}$) when the collector voltage (${{\text{V}}_{{\text{CB}}}}$) which is applied between collector and base is kept constant is known as current amplification factor. It is denoted by $\alpha $.

At constant collector voltage (${{\text{V}}_{{\text{CB}}}}$),
Current amplification factor $\alpha = \dfrac{{\Delta {{\text{I}}_{\text{C}}}}}{{\Delta {{\text{I}}_{\text{E}}}}}$
The ratio of the change in the collector current ($\Delta {{\text{I}}_{\text{C}}}$) to the change in the base current ($\Delta {{\text{I}}_{\text{B}}}$) is known as base current amplification factor. It is denoted by $\beta $.
Current amplification factor $\beta = \dfrac{{\Delta {{\text{I}}_{\text{C}}}}}{{\Delta {{\text{I}}_{\text{B}}}}}$
As we know that the emitter current ${{\text{I}}_{\text{E}}}$ is equal to the base current ${{\text{I}}_{\text{B}}}$ and the collector current ${{\text{I}}_{\text{C}}}$
i.e., ${{\text{I}}_{\text{E}}} = {{\text{I}}_{\text{B}}} + {{\text{I}}_{\text{C}}}$
By taking the change in all the above quantities given in the LHS and RHS, we get
$
  \Delta {{\text{I}}_{\text{E}}} = \Delta {{\text{I}}_{\text{B}}} + \Delta {{\text{I}}_{\text{C}}} \\
   \Rightarrow \Delta {{\text{I}}_{\text{B}}} = \Delta {{\text{I}}_{\text{E}}} - \Delta {{\text{I}}_{\text{C}}} \\
 $
By dividing the above equation by $\Delta {{\text{I}}_{\text{E}}}$ on both the sides, we get
 $
   \Rightarrow \dfrac{{\Delta {{\text{I}}_{\text{B}}}}}{{\Delta {{\text{I}}_{\text{E}}}}} = \dfrac{{\Delta {{\text{I}}_{\text{E}}} - \Delta {{\text{I}}_{\text{C}}}}}{{\Delta {{\text{I}}_{\text{E}}}}} \\
   \Rightarrow \dfrac{{\Delta {{\text{I}}_{\text{B}}}}}{{\Delta {{\text{I}}_{\text{E}}}}} = \dfrac{{\Delta {{\text{I}}_{\text{E}}}}}{{\Delta {{\text{I}}_{\text{E}}}}} - \dfrac{{\Delta {{\text{I}}_{\text{C}}}}}{{\Delta {{\text{I}}_{\text{E}}}}} \\
   \Rightarrow \dfrac{{\Delta {{\text{I}}_{\text{B}}}}}{{\Delta {{\text{I}}_{\text{E}}}}} = 1 - \alpha \\
 $
By dividing the numerator and the denominator by $\Delta {{\text{I}}_{\text{C}}}$ in the LHS of the above equation, we have
\[
 \Rightarrow \dfrac{{\dfrac{{\Delta {{\text{I}}_{\text{B}}}}}{{\Delta {{\text{I}}_{\text{C}}}}}}}{{\dfrac{{\Delta {{\text{I}}_{\text{E}}}}}{{\Delta {{\text{I}}_{\text{C}}}}}}} = 1 - \alpha \\
   \Rightarrow \dfrac{{\dfrac{1}{\beta }}}{{\dfrac{1}{\alpha }}} = 1 - \alpha \\
   \Rightarrow \left( {\dfrac{1}{\beta }} \right)\left( \alpha \right) = 1 - \alpha \\
   \Rightarrow \dfrac{\alpha }{\beta } = 1 - \alpha \\
   \Rightarrow \alpha = \beta \left( {1 - \alpha } \right) \\
   \Rightarrow \alpha = \beta - \alpha \beta \\
   \Rightarrow \alpha + \alpha \beta = \beta \\
   \Rightarrow \alpha \left( {1 + \beta } \right) = \beta \\
   \Rightarrow \alpha = \dfrac{\beta }{{1 + \beta }} \\
 \]
The above equation represents the relation between the current amplification factors $\alpha $ and $\beta $ of a transistor.

Note:
It is very important to note that the equation \[\alpha = \dfrac{\beta }{{1 + \beta }}\] can be written as \[ \Rightarrow \alpha \left( {1 + \beta } \right) = \beta \Rightarrow \alpha + \alpha \beta = \beta \Rightarrow \alpha = \beta - \alpha \beta \Rightarrow \alpha = \beta \left( {1 - \alpha } \right) \Rightarrow \beta = \dfrac{\alpha }{{\left( {1 - \alpha } \right)}}\]. From this relation, we can say that if the current amplification factor $\alpha $ approaches 1, $\beta $ will approach infinity.