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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Physics
Chapter
Semiconductors

BITSAT 2015 · Physics · Semiconductors

In CE transistor amplifier, the audio signal voltage across the...

In CE transistor amplifier, the audio signal voltage across the collector resistance of 2kΩ2 \mathrm{k} \Omega is 2V2 \mathrm{ V}. If the base resistance is 1kΩ1 \mathrm{k} \Omega and the current amplification of the transistor is 100,100, the input signal voltage is

Options

  1. A2mV2 \mathrm{mV}
  2. B3mV3 \mathrm{mV}
  3. C10mV10 \mathrm{mV}
  4. D0.1mV0.1 \mathrm{mV}

Correct answer

C. 10mV10 \mathrm{mV}

Step-by-step solution

Given : Voltage across the collector V0=2V;\mathrm{V}_{0}=2\mathrm{V} ; collector resistance, Rc=2×103Ω;\mathrm{R}_{\mathrm{c}}=2 \times 10^{3} \Omega ; Base resistance RB=1×103Ω;\mathrm{R}_{\mathrm{B}}=1 \times 10^{3} \Omega ; Input signal voltage, Vi=?\mathrm{V}_{\mathrm{i}}=?V0=ICRC=2⇒IC=22×103=10−3A Current gain α=ICIB=100⇒IB=IC100=10−3100=10−5AVi=RBIB⇒Vi=1×103×10−5⇒Vi=10−2V⇒Vi=10mV\begin{array}{l}\mathrm{V}_{0}=\mathrm{I}_{\mathrm{C}} \mathrm{R}_{\mathrm{C}}=2 \\ \Rightarrow \quad \mathrm{I}_{\mathrm{C}}=\frac{2}{2 \times 10^{3}}=10^{-3} \mathrm{ A}\\ \text{ Current gain } \alpha=\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{B}}}=100 \\ \Rightarrow \quad \mathrm{I}_{\mathrm{B}}=\frac{\mathrm{I}_{\mathrm{C}}}{100}=\frac{10^{-3}}{100}=10^{-5} \mathrm{ A}\\ \quad \mathrm{ V}_{\mathrm{i}}=\mathrm{R}_{\mathrm{B}} \mathrm{I}_{\mathrm{B}} \Rightarrow \mathrm{V}_{\mathrm{i}}=1 \times10^{3} \times10^{-5} \\ \Rightarrow \quad \mathrm{V}_{\mathrm{i}}=10^{-2} \mathrm{ V} \Rightarrow \mathrm{V}_{\mathrm{i}}=10 \mathrm{mV} \end{array}$$
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