Question

In a diode AM detector, the output circuit consists of `R = 1 k Omega and C = 10p F.` A carrier signal of 100 kHz is to be detected. Is it good? If yes, then explain why? If not, what value of C would you suggest ?

Answer 1

Correct Answer - `C = 1muF = 10^(-6)F`
Here, `R = 1 k Omega = 10^3 Omega, C = 10pF`
`= 10 xx 10^(-12) F = 10^(-11)F`
`:. RC = 10^3 xx 10^(-11)s = 10^(-8)s and 1/v_c = 1/(100 xx 10^3) s = 10^(-5)s`
We find that `1/v_c` is not less than RC, as is required for
demodulation. Therefore, the arrangement is NO GOOD.
For a satisfactory arrangement, let us try `C = 1muF = 10^(-6)F`
`:. RC = 10^3 xx 10^(-6) s = 10^(-3)s `
Now, `1/v_c = (=10^(-5)s)lt lt RC(=10^(-3)s)`
`:.` The condition is satisfied. This is good enough for demodulation.