Question

Discuss the application of diode as clipper with ideal diode and practical germanium diode. Draw the neat circuit diagrams and waveforms.

Answer 1

Answer:

A clipping circuit or a clipper is a device used to ‘clip’ the input voltage to prevent it from attaining a value larger than a predefined one. As you can see in the picture below this device cuts off the positive or negative peak value of a cycle.

The basic components required for a clipping circuit are – an ideal diode and a resistor. In order to fix the clipping level to the desired amount, a dc battery must also be included. When the diode is forward biased, it acts as a closed switch, and when it is reverse biased, it acts as an open switch. Different levels of clipping can be obtained by varying the amount of voltage of the battery and also interchanging the positions of the diode and resistor.

Depending on the features of the diode, the positive or negative region of the input signal is “clipped” off and accordingly the diode clippers may be positive or negative clippers.

There are two general categories of clippers: series and parallel (or shunt). The series configuration is defined as one where a diode is in series with the load, while the shunt clipper has the diode in a branch parallel to the load.

1. Positive Clipper and Negative Clipper

Positive Diode Clipper

In a positive clipper, the positive half cycles of the input voltage will be removed. The circuit arrangements for a positive clipper are illustrated in the figure given below.

As shown in the figure, the diode is kept in series with the load. During the positive half cycle of the input waveform, the diode ‘D’ is reverse biased, which maintains the output voltage at 0 Volts. This causes the positive half cycle to be clipped off. Dur­ing the negative half cycle of the input, the diode is forward biased and so the nega­tive half cycle appears across the output.

In Figure (b), the diode is kept in parallel with the load. This is the diagram of a positive shunt clipper circuit. During the positive half cycle, the diode ‘D’ is forward biased and the diode acts as a closed switch. This causes the diode to conduct heavily. This causes the voltage drop across the diode or across the load resistance RL to be zero. Thus output voltage during the positive half cycles is zero, as shown in the output waveform. During the negative half cycles of the input signal voltage, the diode D is reverse biased and behaves as an open switch. Consequently, the entire input voltage appears across the diode or across the load resistance RL if R is much smaller than RL.