Question

A half wave rectifier circuit is also known as a clipper. A clipper circuit removes either the positive half or negative half or a part of the input positive or negative cycles of the input AC signal. 

In a typical half wave rectifier circuit, as shown in the circuit diagram, only the positive half cycles are completely removed. This circuit is also known as the Series Positive Clipper.

Similarly, just by reversing the diode, only the negative half cycles are removed. This is Series Negative Clipper circuit.

Now a Series Positive Clipper circuit can be modified by including a battery as shown here.

During the positive input half cycle, the diode is reverse biased by the input supply voltage V_i. But since the positive terminal of the battery is connected to p-side and the negative terminal of the battery is connected to n-side of the diode, the diode is forward biased by the battery voltage V_B. That means the diode is reverse biased by the input supply voltage (V_i) and forward biased by the battery voltage (V_B). Initially, the input supply voltage V_i < V_B , so V_B dominates over the V_i . Hence, the diode is forward biased by the V_B and allows electric current through it. As a result, the signal appears at the output.

As soon as V_i becomes greater than V_B, the diode is reverse biased. So no current flows through the diode. As a result, input signal does not appear at the output. Instead a constant voltage = V_B appears across the output. So the clipping takes place during the positive half cycle, at the time the input V_i is greater than the V_B.

Now consider the negative input half cycle. The diode is forward biased by both V_B and V_i. That means, during the negative half cycle, it doesn’t matter whether the Vi is greater or less than the V_B, the diode remains forward biased. So the complete negative half cycle appears at the output. Thus, the series positive clipper with positive bias clips a small portion of the positive half cycles in the output. 

Answer the following questions:

(a) Draw the circuit diagram to represent the series clipper circuit with a negative bias. 

(b) Which part of the input ac cycle will get clipped in the case of series clipper circuit with a negative bias? Represent the clipped output waveform through this circuit. 

(c) Explain the output waveform during the positive input cycle in the series clipper circuit with a negative bias. 

OR 

Explain the output waveform during the negative input cycle in the series clipper circuit with a negative bias.

Answer 1

(a) 

(b) The negative cycle of the input signal gets clipped.

(c) During the positive half cycle, the diode D is forward biased by both input supply voltage V_i and the battery voltage V_B.

So it doesn’t matter whether the input supply voltage is greater or less than battery voltage V_B, the diode always remains forward biased. Therefore, during the positive half cycle, the signal appears at the output.

OR 

During the negative half cycle, the diode D is reverse biased by the input supply voltage V_i and forward biased by the battery voltage V_B.

Initially, the input supply voltage V_i is less than the battery voltage V_B. So the diode is forward biased by the battery voltage V_B. As a result, the signal appears at the output. When the input supply voltage V_i becomes greater than the battery voltage V_B, the diode will become reverse biased. As a result, no signal appears at the output.