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Assuming the ideal diode, draw the output waveform for the circuit given in figure. Explain the waveform. |
Answer» Solution :Here, `V=20sin(omegat)` and so maximum voltage is `V_(m)=20V`. It means that input voltage would change from +20V to -20V.  (i) For time intervals from 0 to `t_(1)` and from `t_(2)` to `(T)/(2)` when `v_(i) lt 5V`, potential at anode is less than potential at cathode `(V_(A) lt V_(K))` and so diode is reverse biased and so no current would pass through it. Hence input signal voltage would appear directly across load resistance `R_(L)` in the output. Hence WAVEFORM of output voltage `v_(0)` across `R_(L)` would be similar to input voltage `v_(i)`, which is shown in figure 3. (ii) At INSTANTS `t=t_(1) and t=t_(2)`, when `v_(i)=5V`, no current passes through resistance R and diode D and so `v_(0)=v_(i)=5V` which is shown in figure 3. (iii) Now, for time interval from `t_(1)` to `t_(2), v_(i) gt 5V` and so diode is forward biased where its resistance becomes zero and so potential DIFFERENCE across it will also be zero. Hence, in the time interval from `t_(1)` to `t_(2), v_(0)=5` V = constant (which is the battery voltage). (iv) Now, for negative half cycle from `(T)/(2)` to T, diode D is reverse biased and so no current passes through it DUE to infinite resistance and so input signal would pass directly through `R_(L)` and so waveform of `v_(0)` across `R_(L)` would be similar to waveform of `v_(i)` which is shown in figure 3. |
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