The relation between α and β in a transistor
A. <span class="math-tex">\(\alpha = \frac{{1 - \beta }}{\beta }\)</span>
B. <span class="math-tex">\(\alpha = \frac{\beta }{{1 - \beta }}\)</span>
C. <span class="math-tex">\(\alpha = \frac{\beta }{{1 + \beta }}\)</span>
D. <span class="math-tex">\(\alpha = \frac{{1 + \beta }}{\beta }\)</span>
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Right Answer is: C
SOLUTION
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'
\(β = \frac{α }{{1-α}}\)