question_answer 54)

  Assuming the ideal diode, draw the output waveform for the circuit given in Fig. 9(EP).18(a). Explain the waveform.

Answer:

                  When the input voltage is equal to or less than 5 V, diode will be reverse biased. It will offer high resistance in comparison to resistance (R) in series. Now diode appears in open circuit. The input waveform is then passed to the output terminals. The result with sine wave input is to clip off all positive-going portion above + 5V volt.                                                            If input voltage is more than + 5 V, diode will be conducting as if forward biased offering low resistance incomparison to R. But there will be no voltage in output beyond 5 volt as the voltage beyond + 5 V will appear across R. When input voltage is negative, there will be opposition to 5 V battery in p-n junction circuit. Due to it, reverse bias voltage of p-n junction decreases and a voltage appears across output. When input voltage becomes more than - 5 V, the diode will be reverse biased. It will offer high resistance in comparison to resistance R in series. Now junction diode appears in open circuit. The input wave form is then passed on to the output terminals. The output wave form will be as shown in Pig. 9(EP).18(b).