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BITSAT 2022
Exam
BITSAT
Year
2022
Subject
Physics
Chapter
Semiconductors

BITSAT 2022 · Physics · Semiconductors

A transistor is connected in common-emitter (CE) configuration. The...

A transistor is connected in common-emitter (CE) configuration. The collector supply is 8V8 \mathrm{ V} and the voltage drop across a resistor is 500Ω500 \Omega in the collector circuit is 0.6V0.6 \mathrm{ V}. If the current gain factor α\alpha is 0.960.96, find the base current

Options

  1. A25μA25 \mu \mathrm{A}
  2. B50μA50 \mu \mathrm{A}
  3. C20μA20 \mu \mathrm{A}
  4. D35μA35 \mu \mathrm{A}

Correct answer

B. 50μA50 \mu \mathrm{A}

Step-by-step solution

Given, VCC=8VV_{C C}=8 \mathrm{ V} Voltage drop, V0=ΔICRL=0.6VV_0=\Delta I_C R_L=0.6 \mathrm{ V} Load Resistor, RL=500ΩR_L=500 \Omega Gain factor, α=0.96∵β=α1−α=0.961−0.96=24\alpha=0.96\because \quad \beta=\frac{\alpha}{1-\alpha}=\frac{0.96}{1-0.96}=24 Now, V0=ΔICRL0.6=ΔIC×500ΔIC=0.6500=1.2×10−3A\quad V_0=\Delta I_C R_L0.6=\Delta I_C \times 500\Delta I_C=\frac{0.6}{500}=1.2 \times 10^{-3} \mathrm{ A} Now, β=ΔICΔIB∴\beta=\frac{\Delta I_C}{\Delta I_B}\therefore Base current ΔIB=ΔICβ=1.2×10−324=5×10−5=50×10−6ΔIB=50μA\Delta I_B=\frac{\Delta I_C}{\beta}=\frac{1.2 \times 10^{-3}}{24}=5 \times 10^{-5}=50 \times 10^{-6}\Delta I_B=50 \mu \mathrm{A}
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