Question

In a diode Am detector, the output circuit consists of \[R = 1k\Omega \] and \[C = pF\] . A carrier signal of \[100kHz\]is to be detected. Is it good ?
A) Yes
B) No
C) Information is not sufficient
D) None of the above

Answer 1

Hint: The diode detector is the simplest and most basic form of amplitude modulation, AM signal detector and it detects the envelope of the AM signal. The AM diode detector is an envelope detector that provides an output of the envelope of the signal. As such the diode detector or demodulator is able to provide an output proportional to the amplitude of the envelope of the amplitude modulated signal.

Complete step by step answer:
A number of methods can be used to demodulate AM, but the simplest is a diode detector. It operates by detecting the envelope of the incoming signal which it does by rectifying the signal. Current is allowed to flow through the diode in only one direction, giving either the positive or negative half of the envelope at the output.
Therefore, for better demodulation
\[\dfrac{1}{{{f_c}}} < < RC\]
Given:
\[R = 1k\Omega \]
\[C = pF\]
On simplifying the values we have
\[\dfrac{1}{{{f_c}}} = \dfrac{1}{{100 \times {{10}^3}}} = {10^{ - 5}}s\]
\[RC = {10^{}} \times 10 \times {10^{ - 12}} = {10^{ - 8}}s\]
As\[p = {10^{ - 12}}\]
From the above result we conclude that,
Clearly \[\dfrac{1}{{{f_c}}}\] is not less than RC
Therefore, It is not good and option B is correct.

Note: The AM detector or demodulator includes a capacitor at the output. Its purpose is to remove any radio frequency components of the signal at the output. The value is chosen so that it does not affect the audio base-band signal. There is also a leakage path to enable the capacitor to discharge, but this may be provided by the circuit into which the demodulator is connected.