JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
In the figure, given that \(V_{BB}\) supply can vary from \(0\) to \(5.0 \,V,\,\,V_{CC} = 5\,V,\) \(\beta _{dc} = 200,\,\,R_B = 100,\,k\Omega ,\) \(R_C = 1\,k\Omega \) an \(V_{BE} = 1.0\,V,\) The minimum base current and the input voltage at which the transistor will go to saturation, will be respectively
- A \(25\,\mu A\) and \(3.5\,\,V\)
- B \(20\,\mu A\) and \(3.5\,\,V\)
- C \(25\,\mu A\) and \(2.8\,\,V\)
- D \(20\,\mu A\) and \(2.8\,\,V\)
Answer & Solution
Correct Answer
(A) \(25\,\mu A\) and \(3.5\,\,V\)
Step-by-step Solution
Detailed explanation
When switched on : \(\mathrm{V}_{\mathrm{CE}}=0\) \(\mathrm{V}_{\mathrm{CC}}-\mathrm{R}_{\mathrm{C}} \mathrm{I}_{\mathrm{C}}=0\) \(\mathrm{i}_{\mathrm{C}}=\frac{\mathrm{V}_{\mathrm{CC}}}{\mathrm{R}_{\mathrm{C}}}\) \(=5 \times 10^{-3} \mathrm{A}\) \(I_C\,=\,Bi_B\)…
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