Determine the peak value of the output signal of the half wave rectifier or a series clipper realised using a single diode having forward voltage of 0.6 V, if the input applied the circuit is periodic sinusoidal signal of 10 V peak to peak voltage. 

Explanation:

Half-Wave Rectifier or Series Clipper Using a Single Diode

Definition: A half-wave rectifier is an electronic circuit that allows only one half of the input alternating current (AC) signal to pass through, effectively clipping the other half. This is achieved using a single diode, which conducts during the positive half-cycle of the AC signal and blocks during the negative half-cycle. A series clipper, similarly, is a circuit designed to clip (restrict) a portion of the input signal based on the diode's characteristics and configuration.

Problem Statement: Determine the peak value of the output signal of the half-wave rectifier or series clipper realized using a single diode having a forward voltage of 0.6 V, if the input applied to the circuit is a periodic sinusoidal signal of 10 V peak-to-peak voltage.

Solution:

To solve this problem, let us analyze the circuit step by step:

1. Input Signal Characteristics:

• The input is a sinusoidal signal with a peak-to-peak voltage of 10 V.
• The peak-to-peak voltage indicates the total voltage swing from the positive peak to the negative peak.
• For a sinusoidal signal, the peak voltage (V_peak) is half of the peak-to-peak voltage:

V_peak = (Peak-to-Peak Voltage) ÷ 2 = 10 ÷ 2 = 5 V

This means the input signal swings from +5 V (positive peak) to -5 V (negative peak).

2. Diode Characteristics:

• The diode used in the circuit has a forward voltage drop of 0.6 V. This is the minimum voltage required for the diode to conduct.
• When the input voltage exceeds 0.6 V (during the positive half-cycle), the diode will conduct, allowing the signal to pass through.
• During the negative half-cycle, the diode will block the signal, and the output voltage will be zero.

3. Output Signal Analysis:

• During the positive half-cycle of the input signal:
• The diode conducts when the input voltage exceeds 0.6 V.
• The output voltage is equal to the input voltage minus the forward voltage drop of the diode.
• At the peak of the input signal (+5 V), the output voltage is:

V_output = V_input - V_forward = 5 V - 0.6 V = 4.4 V

• During the negative half-cycle of the input signal:
• The diode is reverse-biased and does not conduct.
• The output voltage is zero.

4. Peak Value of the Output Signal:

The peak value of the output signal is the maximum voltage achieved during the positive half-cycle, which is:

V_{peak, output} = 4.4 V

Correct Answer: Option 4

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: 5 V

This option assumes that the output voltage is equal to the peak voltage of the input signal. However, it fails to account for the diode's forward voltage drop (0.6 V). The correct peak output voltage is 4.4 V, not 5 V.

Option 2: 9.4 V

This option is incorrect because it exceeds the peak voltage of the input signal. The input signal has a peak voltage of 5 V, so the output signal cannot have a peak voltage of 9.4 V.

Option 3: 10 V

This option is incorrect because it corresponds to the peak-to-peak voltage of the input signal, not the peak output voltage. The diode's forward voltage drop further reduces the output peak voltage to 4.4 V.

Option 4: 4.4 V

This is the correct option as it accurately accounts for the input signal's peak voltage and the diode's forward voltage drop.

Conclusion:

A half-wave rectifier or series clipper using a single diode clips the negative half-cycle of the input signal and reduces the positive half-cycle by the diode's forward voltage drop. For an input sinusoidal signal with a peak-to-peak voltage of 10 V and a diode with a forward voltage of 0.6 V, the peak value of the output signal is 4.4 V.