Correct Answer - Option 1 : β = α/(1 - α)
The transistor α and β are related as:
\(α = \frac{β }{{β + 1}}\)
This can also be written as:
\(β = \frac{α }{{1-α}}\)
Derivation:
The common-emitter current gain (β) is the ratio of the transistor's collector current to the transistor's base current, i.e.
\(β = \frac{{{I_C}}}{{{I_B}}}\)
And the common base DC current gain (α) is a ratio of the transistor's collector current to the transistor's emitter current, i.e.
\(α = \frac{{{I_C}}}{{{I_E}}}\)
The transistor currents are related by the relation:
IE = IB + IC
α can now be written as:
\(α = \frac{{{I_C}}}{{{I_B+I_C}}}\)
Dividing both the numerator and denominator by IB, we get:
\(α = \frac{{{I_C/I_B}}}{{{1+I_C/I_B}}}\)
Since \(β = \frac{{{I_C}}}{{{I_B}}}\)
\(α = \frac{β }{{β + 1}}\) 'or'
\(β = \frac{α }{{1-α}}\)