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Question: In a diode AM detector with the output circuit consists of R = 1 M\(\Omega\) and C = 1 pF would be m...

In a diode AM detector with the output circuit consists of R = 1 MΩ\Omega and C = 1 pF would be more suitable for detecting a carrier signal of

A

1 MHz

B

0.1 MHz

C

0.5 MHz

D

10 MHz

Answer

10 MHz

Explanation

Solution

: Here, RC=106×1012=106s= 10^{6} \times 10^{- 12} = 10^{- 6}s

For demodulation, 1υc<<RC\frac{1}{\upsilon_{c}} < < RC

υc>>1RCorυc>>1106=106Hz\upsilon_{c} > > \frac{1}{RC}or\upsilon_{c} > > \frac{1}{10^{- 6}} = 10^{6}Hz

υc>>106Hz\therefore\upsilon_{c} > > 10^{6}Hz