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The load on an `Am` diode detector consists of a resistance of `50KOmega` in parallel with a capacitor of `0.001 muF`. Determine the maximum modulation index that the detector can handle without distortion when modulation frequency is (`i`) `1kHz` (`ii`) `5 kHz` |
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Answer» Here `R_(C)` and `C` are in parallel combination and phase difference between `i_(1)` and `i_(2)` is `(pi)/(2)` `i^(2)+i_(1)^(2)+i_(2)^(2)` `[(V)/(Z_(m))]^(2)=[(V)/(R_(c))]^(2)+[(V)/(X_(c))]^(2)` `Z_(m)=(1)/(sqrt((1)/(R_(c))^(2)+(1)/((X_(c))^(2))))` `M_(max)=(Z_(m))/(R_(c))` `=(1)/(R_(c)sqrt((1)/(R_(c))^(2)+(1)/((X_(c))^(2))))` `=(1)/(sqrt(1+(2pifCR_(c))^(2)))` For `f=1kHz` `M_(max)=(1)/(sqrt(1+0.098696))=0.945` For `f=5kHz` `M_(max) =(1)/(sqrt(1+24674))=0.537` |
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