Question

Assuming the ideal diode, draw the output waveform for the circuit given in figure. Explain the waveform

Answer 1

When signal $20\sin \omega t$ gives input voltage less than $5$ volt (because after $5V$ diode will get positive voltage at its P-junction) then diode will be in reverse bias so resistance of diode remain infinity so input signal will not pass through diode and battery path so output across $A$ and $B$ will increase from $0-5V$ (graph OC).
Now when the input voltage $20\sin \omega t$ increase beyond $5V$ then path of diode $5V$ battery will offer very low resistance, so the current passes through diode and battery and output (across $A$ and $B$) remain $5V$ (graph CD).
Now when the voltage decreases the diode will be in reverse bias and output will again fall from $5V$ to $0V$ as input changes (graph ED). When input voltage become negative (there is opposite of $5V$ battery in p-n junction input voltage becomes more than $5V$ now) the diode is in reversed bias it will not conduct current through $CD$, and in output across $AB$ will get same as input $AC$ ie., for negaive cycle diode offer infinite resistance as compared to $R$ in series graph $E,F,G$
Same repetition of input and output continues-graph showing the output waveform .