JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
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
- A \(I _b=1.0\,\mu A\)
- B \(I _{ b }=0.10\,\mu A\)
- C \(I_b=100\,\mu A\)
- D \(I _{ b }=10\,\mu A\)
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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