IC = [a / (1 - a )] IB +_____________?
Correct answer: A. ICEO
- A. ICEO
- B. ICBO
- C. IC
- D. (1 - a ) IB
Explanation
Starting with I_C = βI_B + I_CEO and β = α/(1 − α), the relation becomes I_C = [α/(1 − α)]I_B + I_CEO. Here I_CEO is the common-emitter leakage current, already including the amplified leakage component.
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