Welcome to the NCERT Solutions for Class 12 Physics - Chapter Semiconductor Electronics. This page offers a step-by-step solution to the specific question from Exercise 1, Question 9: . With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.
For a CE-transistor amplifier, the audio signal voltage across the collected resistance of 2 kΩ is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ.
Collector resistance, R C = 2 kΩ = 2000 Ω
Audio signal voltage across the collector resistance, V = 2 V
Current amplification factor of the transistor, β = 100
Base resistance, R B = 1 kΩ = 1000 Ω
Input signal voltage = V i
Base current = I B
We have the amplification relation as: V/Vi = β Rc/Rb
Voltage amplification, Vi = V Rb / β Rc
= 2x1000 / 100x2000 = 0.01 V
Therefore, the input signal voltage of the amplifier is 0.01 V.
Base resistance is given by the relation: RB = Vi / IB
= 0.01 / 1000 = 10 μA
Therefore, the base current of the amplifier is 10 μA.
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Read MoreWelcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 9: For a CE-transistor amplifier, the audio signal voltage across the collected resistance of 2 k&Omega....
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