Saved Bookmarks
| 1. |
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» 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. |
|