JEE MAIN - Physics (2013 (Offline) - No. 6)
A diode detector is used to detect an amplitude modulated wave of $$60\% $$ modulation by using a condenser of capacity $$250$$ picofarad in parallel with a load resistance $$100$$ kilo $$ohm.$$ Find the maximum modulated frequency which could be detected by it.
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$$10.62$$ $$MHz$$
$$10.62$$ $$kHz$$
$$5.31$$ $$MHz$$
$$5.31$$ $$kHz$$
Explanation
Given : Resistance $$R=100$$ kilo $$ohm=100$$ $$ \times {10^3}\Omega $$
$$\therefore$$ Capacitance $$C=250$$ picofarad $$ = 250 \times {10^{ - 12}}F$$
$$\tau = RC = 100 \times {10^3} \times 250 \times {10^{ - 12}}\,\sec $$
$$ = 2.5 \times {10^7} \times {10^{ - 12}}\sec $$
$$ = 2.5 \times {10^{ - 5}}\sec $$
The higher frequency which can be detected with tolerable distortion is
$$f = {1 \over {2\pi {m_a}RC}}$$
$$ = {1 \over {2\pi \times 0.6 \times 2.5 \times {{10}^{ - 5}}}}Hz$$
$$ = {{100 \times {{10}^4}} \over {2.5 \times 1.2\pi }}Hz$$
$$ = 10.61KHz$$
This condition is obtained by applying the condition that rate of decay of capacitor voltage must be equal or less than the rate of decay modulated signal voltage for proper detection of modulated signal.
$$\therefore$$ Capacitance $$C=250$$ picofarad $$ = 250 \times {10^{ - 12}}F$$
$$\tau = RC = 100 \times {10^3} \times 250 \times {10^{ - 12}}\,\sec $$
$$ = 2.5 \times {10^7} \times {10^{ - 12}}\sec $$
$$ = 2.5 \times {10^{ - 5}}\sec $$
The higher frequency which can be detected with tolerable distortion is
$$f = {1 \over {2\pi {m_a}RC}}$$
$$ = {1 \over {2\pi \times 0.6 \times 2.5 \times {{10}^{ - 5}}}}Hz$$
$$ = {{100 \times {{10}^4}} \over {2.5 \times 1.2\pi }}Hz$$
$$ = 10.61KHz$$
This condition is obtained by applying the condition that rate of decay of capacitor voltage must be equal or less than the rate of decay modulated signal voltage for proper detection of modulated signal.
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