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JEE Main 20137 April
Exam
JEE Main
Year
2013
Date
7 April
Subject
Physics
Chapter
Semiconductors

JEE Main 2013 · 7 April · Physics · Semiconductors

A diode detector is used to detect an amplitude modulated wave of...

A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 pico farad in parallel with a load resistance of 100 kilo ohm. Find the maximum modulated frequency which could be detected by it.

A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250

Options

  1. A5.31 MHz\text{5.31 MHz}
  2. B5.31 kHz\text{5.31 kHz}
  3. C10.62 MHz\text{10.62 MHz}
  4. D10.62 kHz\text{10.62 kHz}

Correct answer

D. 10.62 kHz\text{10.62 kHz}

Step-by-step solution

A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250

τ=RC=100×103×250×10−12secτ = \text{RC} = 1 0 0 \times 1 {0}^{3} \times 2 5 0 \times 1 {0}^{- 1 2} \text{sec}

=2.5×107×10−12sec= \text{2.5} \times 1 {0}^{7} \times 1 {0}^{-1 2} \text{sec}

=2.5×10−5sec= \text{2.5} \times 1 {0}^{- 5} \text{sec}

The higher frequency which can be detected with tolerable distortion is

f=12πmaRC=12π×0.6×2.5×10−5Hz\text{f} = \dfrac{1}{2 \pi {\text{m}}_{\text{a}} \text{RC}} = \dfrac{1}{2 \pi \times \text{0.6} \times \text{2.5} \times 1 {0}^{- 5}} \text{Hz}

=100×10425×1.2πHz= \dfrac{100 \times 1{0}^{4}}{25 \times \text{1.2}\pi}\text{Hz}

=41.2π×104Hz= \dfrac{4}{\text{1.2} \pi} \times 1 {0}^{4} \text{Hz}

=10.61 kHz= \text{10.61 kHz}

This condition is obtained by applying the condition that rate of decay of capacitor voltage must be equal or less then the rate of decay modulated signal voltage for proper detection of modulated signal.
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