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JEE Mains · Physics · STD 12 - Semicondutor electronics
  • JEE Mains 2023
  • Medium
  • MCQ

For a given transistor amplifier circuit in \(CE\) configuration \(V_{C C}=1 V , R_c=1 k \Omega, R_b=100 k \Omega\) and \(\beta=100\). Value of base current \(I_b\) is

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Answer & Solution

Correct Answer

(D) \(I _{ b }=10\,\mu A\)

Step-by-step Solution

Detailed explanation

Considering the transistor in saturation mode

\(V _{ CE }=0\)

Using KVL

\(-I_c R_c+V_{c c}=0\)

\(I_c=\frac{V_{ cC }}{R_c}=\frac{1}{1 \times 10^3}\)

\(I_c=10^{-3}\,A\)

\(\beta=\frac{I_c}{I_b}\)

\(I_b=\frac{10^{-3}}{100} \Rightarrow 10^{-5}\,A \Rightarrow I_b=10\,\mu A\)

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