In a diode AM detector,the output circuit has | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For proper demodulation in an $AM$ detector,the time constant $RC$ must satisfy the condition: $\frac{1}{f_c} \ll RC < \frac{1}{f_m}$,where $f_c$

Vedclass · Accepted Answer

No

Vedclass · Answer

(B) For proper demodulation in an $AM$ detector,the time constant $RC$ must satisfy the condition: $\frac{1}{f_c} \ll RC First,calculate the period of the carrier signal: $\frac{1}{f_c} = \frac{1}{100 	imes 10^3 \ Hz} = 10^{-5} \ s$.Next,calculate the time constant of the circuit: $RC = (10^3 \ \Omega) 	imes (10 	imes 10^{-12} \ F) = 10^{-8} \ s$.Comparing the two,we see that $RC = 10^{-8} \ s$ is much smaller than $\frac{1}{f_c} = 10^{-5} \ s$.Since the condition $\frac{1}{f_c} \ll RC$ is not satisfied,the circuit cannot effectively detect the $100 \ kHz$ carrier signal.

In a diode $AM$ detector,the output circuit has $R = 1 \ k\Omega$ and $C = 10 \ pF$. Can it detect a carrier signal of $100 \ kHz$?

Similar Questions

If the output frequency of a full-wave rectifier is $100 \text{ Hz}$, then the input frequency will be . . . . . . . (in $\text{ Hz}$)

Choose the correct waveform that can represent the voltage across $R$ of the following circuit,assuming the diode is ideal one:

In a full-wave rectifier circuit, the input frequency is $50 \, Hz$. The fundamental frequency of the ripple is ....... $Hz$.

In the diode-based rectifier circuit given below, if $V_s=V_m \sin \omega t$ and the diode is ideal, then the average value of $V_L$ is

The output current versus time curve of a rectifier is shown in the figure. The average value of output current in this case is

Vedclass Test Series

Exam Paper Generator

Online Exam Module