
In a diode detector, the output circuit consists of $R = 1M\Omega $ and $C = 1pF$. Calculate the carrier frequency, it can detect.
A. $3000Hz$
B. $1MHz$
C. $2000Hz$
D. $4000Hz$
Answer
565.5k+ views
Hint: A diode detector is known as the simplest form of envelope detector. A diode detector is simply defined as a diode that is connected in between the input and output of the circuit in which resistor and capacitor are connected in parallel. The frequency of a diode detector is greater than the inverse of the product of the resistance and capacitance in the circuit.
Formula used:
The formula used for calculating the frequency of diode detector is given by
${\nu _c} \gg \,\dfrac{1}{{RC}}$
Here, ${\nu _c}$ is the frequency of the diode detector, $R$ is the resistance in the circuit and $C$ is the capacitance in the circuit.
Complete step by step answer:
A diode detector is known as the simplest form of envelope detector. A diode detector is simply defined as a diode that is connected in between the input and output of the circuit in which resistor and capacitor are connected in parallel.
As given in the question, the resistance in the circuit, $R = 1M\Omega = \,{10^6}\Omega $
Also, the capacitance in the circuit, $C = 1pF = {10^{ - 12}}F$
The product of the resistance and the capacitance is
$RC = {10^6} \times {10^{ - 12}} = {10^{ - 6}}s$
The formula used for calculating the frequency of the diode detector is given by
${\nu _c} \gg \,\dfrac{1}{{RC}}$
Putting the value of resistance and the capacitance in the above equation we get
${\nu _c} \gg \,\dfrac{1}{{{{10}^{ - 6}}}}$
$ \Rightarrow \,{\nu _c} \gg \,{10^6}Hz$
$ \therefore \,{\nu _c} \gg \,1MHz$
Therefore, the carrier frequency detected by the diode detector is $1MHz$.
Hence, option B is the correct option.
Note:The formula of the frequency of the diode detector is also known as demodulation formula. Demodulation, which is also known as detection, is defined as the process of regenerating original modulating frequency from the radio frequency. This frequency is known as the carrier frequency of the diode detector.
Formula used:
The formula used for calculating the frequency of diode detector is given by
${\nu _c} \gg \,\dfrac{1}{{RC}}$
Here, ${\nu _c}$ is the frequency of the diode detector, $R$ is the resistance in the circuit and $C$ is the capacitance in the circuit.
Complete step by step answer:
A diode detector is known as the simplest form of envelope detector. A diode detector is simply defined as a diode that is connected in between the input and output of the circuit in which resistor and capacitor are connected in parallel.
As given in the question, the resistance in the circuit, $R = 1M\Omega = \,{10^6}\Omega $
Also, the capacitance in the circuit, $C = 1pF = {10^{ - 12}}F$
The product of the resistance and the capacitance is
$RC = {10^6} \times {10^{ - 12}} = {10^{ - 6}}s$
The formula used for calculating the frequency of the diode detector is given by
${\nu _c} \gg \,\dfrac{1}{{RC}}$
Putting the value of resistance and the capacitance in the above equation we get
${\nu _c} \gg \,\dfrac{1}{{{{10}^{ - 6}}}}$
$ \Rightarrow \,{\nu _c} \gg \,{10^6}Hz$
$ \therefore \,{\nu _c} \gg \,1MHz$
Therefore, the carrier frequency detected by the diode detector is $1MHz$.
Hence, option B is the correct option.
Note:The formula of the frequency of the diode detector is also known as demodulation formula. Demodulation, which is also known as detection, is defined as the process of regenerating original modulating frequency from the radio frequency. This frequency is known as the carrier frequency of the diode detector.
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