JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
An \(NPN\) transistor is used in common emitter configuration as an amplifier with \(1\,k\Omega \) load resistance. Signal voltage of \(10\,mV\) is applied across the base emitter. This produces a \(3\,mA\) change in the collector current and \(15\,\mu A\) change in the base current of the amplifier. The input resistance and voltage gain are
- A \(0.67\,k\Omega ,300\)
- B \(0.67\,k\Omega ,200\)
- C \(0.33\,k\Omega ,1.5\)
- D \(0.33\,k\Omega ,300\)
Answer & Solution
Correct Answer
(A) \(0.67\,k\Omega ,300\)
Step-by-step Solution
Detailed explanation
Input current \(=15 \times 10^{-6}\) Output current \(=3 \times 10^{-3}\) Resistance out put \(=1000\) \(\mathrm{V}_{\text {input }}=10 \times 10^{-3}\) Now \(V_{input}\) \(=\) \(r_{input}\) \(\times\) \(i_{input}\)…
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