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JEE Mains · Physics · STD 12 - Semicondutor electronics
  • JEE Mains 2019
  • Medium
  • Numerical

A common emitter amplifier circuit, built using an \(npn\) transistor, is shown in the figure. Its \(dc\) current gain is \(250\), \(R_C = 1\,k\, \Omega \) and \(V_{CC} = 10\,V\). What is the minimum base current for \(V_{CE}\) to reach saturation?.........\(\mu A\)

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Answer & Solution

Correct Answer

(B) \(40\)

Step-by-step Solution

Detailed explanation

At saturation state, \(V_{CE}\) becomes zero

\(\Rightarrow \mathrm{i}_{\mathrm{C}}\) \(=\frac{10 \mathrm{V}}{1000 \Omega}=10 \mathrm{mA}\)

Now current gain factor \(\beta=\frac{\mathrm{i}_{\mathrm{C}}}{\mathrm{i}_{\mathrm{B}}}\)

\(\Rightarrow \mathrm{i}_{B}=\frac{10 \mathrm{mA}}{250}=40 \mu \mathrm{A}\)

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