Question
Download Solution PDFIn a common base configuration, the alpha of the transistor is 0.99, its collector current is 1 mA and the collector to base current with emitter open is 1 μA. The value of base current is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
In a common base connection,
IE = IB + IC
And IC = α IE + ICBO
Where α is the current amplification factor
\(I_C= \frac{α}{1-α} I_B+\frac {1}{1-α}I_{CBO}\)
Calculation:
Given that,
α = .99
ICBO = 1 μA
IC = 1 mA
⇒ \(I_C= \frac{α}{1-α} I_B+\frac {1}{1-α}I_{CBO}\)
⇒ \(1 mA= \frac{.99}{1-.99} I_B+\frac {1}{1-.99} \times 1 μ A\)
⇒ IB ≈ 9 μA
Last updated on Jun 23, 2025
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