TS EAMCET · Physics · Semiconductors
The current amplification factor of a transistor in common emitter configuration is 80. If the emitter current is 2.43 mA, then the base current is
- A \(15 \mu \mathrm{~A}\)
- B \(1.5 \mu \mathrm{~A}\)
- C \(3 \mu \mathrm{~A}\)
- D \(30 \mu \mathrm{~A}\)
Answer & Solution
Correct Answer
(D) \(30 \mu \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
\(I_b = \frac{I_e}{\beta + 1}\) \(I_b = \frac{2.43 \times 10^{-3} \text{ A}}{80 + 1}\) \(I_b = \frac{2.43 \times 10^{-3} \text{ A}}{81}\) \(I_b = 30 \times 10^{-6} \text{ A} = 30 \mu \text{A}\)
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