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A receiver is operated at a temperature of 300 K. The transistor used in the receiver have an average output resistance of 1 k. The Johnson noise voltage for a receiver with a bandwidth of 200 kHz is
  • a)
    1.8 µV
  • b)
    8.4 µV
  • c)
    4.3 µV
  • d)
    12.6 µV
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A receiver is operated at a temperature of 300 K. The transistor used ...
v2 = 4kBTR where symbols have their usual meanings, hence v = 1.8 µV.
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A receiver is operated at a temperature of 300 K. The transistor used ...
The Johnson noise voltage (Vn) can be calculated using the following formula:

Vn = sqrt(4 * k * T * R * B)

Where:
- Vn is the Johnson noise voltage
- k is the Boltzmann constant (1.38 × 10^-23 J/K)
- T is the temperature in Kelvin (300 K)
- R is the resistance in ohms (1 kΩ)
- B is the bandwidth in hertz (200 kHz = 200,000 Hz)

Plugging in the values:

Vn = sqrt(4 * (1.38 × 10^-23 J/K) * (300 K) * (1,000 Ω) * (200,000 Hz))

Vn ≈ 1.8 x 10^-8 volts

So the Johnson noise voltage for the receiver with a bandwidth of 200 kHz is approximately 1.8 x 10^-8 volts.
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A receiver is operated at a temperature of 300 K. The transistor used in the receiver have an average output resistance of 1 k. The Johnson noise voltage for a receiver with a bandwidth of 200 kHz isa)1.8 µVb)8.4 µVc)4.3 µVd)12.6 µVCorrect answer is option 'A'. Can you explain this answer?
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A receiver is operated at a temperature of 300 K. The transistor used in the receiver have an average output resistance of 1 k. The Johnson noise voltage for a receiver with a bandwidth of 200 kHz isa)1.8 µVb)8.4 µVc)4.3 µVd)12.6 µVCorrect answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A receiver is operated at a temperature of 300 K. The transistor used in the receiver have an average output resistance of 1 k. The Johnson noise voltage for a receiver with a bandwidth of 200 kHz isa)1.8 µVb)8.4 µVc)4.3 µVd)12.6 µVCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A receiver is operated at a temperature of 300 K. The transistor used in the receiver have an average output resistance of 1 k. The Johnson noise voltage for a receiver with a bandwidth of 200 kHz isa)1.8 µVb)8.4 µVc)4.3 µVd)12.6 µVCorrect answer is option 'A'. Can you explain this answer?.
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