Q.17 Define the current gain \(\alpha_{DC}\) and \(\beta_{DC}\) for a transistor. Obtain the relation between them.
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Answer: \(\alpha_{DC} = \dfrac{I_C}{I_E}\), \(\beta_{DC} = \dfrac{I_C}{I_B}\); \(\beta_{DC} = \dfrac{\alpha_{DC}}{1 - \alpha_{DC}}\) and \(\alpha_{DC} = \dfrac{\beta_{DC}}{1 + \beta_{DC}}\).
\(\alpha_{DC}\) (common base d.c. current gain) is the ratio of the collector current to the emitter current: \(\alpha_{DC} = \dfrac{I_C}{I_E}\). Its value is slightly less than 1.
\(\beta_{DC}\) (common emitter d.c. current gain) is the ratio of the collector current to the base current: \(\beta_{DC} = \dfrac{I_C}{I_B}\). It is large (typically 50 to 300).
Relation: In a transistor, \(I_E = I_B + I_C\).
Dividing by \(I_C\): \(\dfrac{I_E}{I_C} = \dfrac{I_B}{I_C} + 1\), i.e. \(\dfrac{1}{\alpha_{DC}} = \dfrac{1}{\beta_{DC}} + 1\).
So \(\dfrac{1}{\beta_{DC}} = \dfrac{1}{\alpha_{DC}} - 1 = \dfrac{1 - \alpha_{DC}}{\alpha_{DC}}\), giving \(\beta_{DC} = \dfrac{\alpha_{DC}}{1 - \alpha_{DC}}\).
Similarly, \(\dfrac{1}{\alpha_{DC}} = \dfrac{1 + \beta_{DC}}{\beta_{DC}}\), so \(\alpha_{DC} = \dfrac{\beta_{DC}}{1 + \beta_{DC}}\).
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