Question

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

Answer 1

An ideal diode is a diode which it has very large resistance in reverse biased and very, low resistance in forward biased. So, it acts like a perfect conductor when voltage is applied forward-biased and like a perfect insulator when voltage is applied reverse-biased.

In reverse biased when the input voltage is equal to or less than 5 V diode, then it will offer high resistance in comparison to resistance (R) in series. Now, the diode appears in an open circuit. The input waveform is then passed to the output terminals. The result with sin wave input is to dip off all positive going portions above 5 V.

If the input voltage is greater than +5 V, the diode is in conducting state, then it will be conducting as if forward biased offering low resistance in comparison to R. But there will be no voltage in output beyond 5 V as the voltage beyond +5 V will appear across R.

When the input voltage is negative, there will be opposition to 5 V battery in p-n junction 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 an open circuit. The input waveform is then passed on to the output terminals.

The output waveform will be like this (as shown below).