# Electrical paper | Physics homework help

In this part of the term paper, you will explain why we need rectification and how to build a single-phase bridge rectifier.

Carefully review these detailed instructions for requirements of this term paper.. Use illustrations if necessary to complete the assignment.

1. Half-Wave Rectifier

In half-wave rectification of a single-phase supply, either the positive or negative half of the AC wave is passed, while the other half is blocked. Because only one half of the input waveform reaches the output, the DC output voltage is lower. Half-wave rectification requires a single diode in a single-phase supply. Rectifiers yield a unidirectional but pulsating direct current; half-wave rectifiers produce far more ripple than full-wave rectifiers, and much more filtering is needed to eliminate harmonics of the AC frequency from the output. Following is a Multisim simulation of a half-wave rectifier without a low pass filter.

Figure 1. Half-wave rectifier without a filter

Based on the output waveform, calculate the DC component using equation (5.16)

and the first harmonic of the output using equations (5.17)

and (5.18)                                                                               with h=1.

In order to block the harmonics and allow only the DC component to the load, a low pass filter is needed. This second order low pass filter consists of an inductor and a capacitor, with its corner frequency being well below the first harmonic frequency. The Multisim simulation with the filter included is shown in figure 2 below.

Figure 2. Half-wave rectifier with a low pass filter.

Calculate the corner frequency of the low pass filter to verify that it is much smaller than 60 Hz. Compare the DC output of the rectifier with the low pass filter and compare it to the earlier calculations. Comment on these results.

2. Full-Wave Rectifier

A full-wave rectifier converts the whole of the input waveform to one of constant polarity (positive or negative) at its output. Full-wave rectification converts both polarities of the input waveform to DC (direct current), and yields a higher DC output voltage. Two diodes and a center tapped transformer, or four diodes in a bridge configuration and any AC source (including a transformer without center tap), are needed. Figure 3 shows a bridge rectifier without a low pass filter.

Figure 3. Full-wave rectifier without a filter.

Based on the output waveform, calculate the DC component using equation (5.16) and the second harmonic of the output using equations (5.17) and (5.18) with h=2.

Again, in order to block the harmonics and allow only the DC component to the load, a low pass filter is needed. This second order low pass filter consists of an inductor and a capacitor, with its corner frequency being well below the second harmonic frequency. The Multisim simulation with the filter included is shown in figure 4 below.

Figure 4. Full-wave rectifier with a low pass filter

Again, calculate the corner frequency of the low pass filter to verify that it is much smaller than 120 Hz. Compare the DC output of the rectifier with the low pass filter and compare it to the earlier calculations.

Final question: why does the full-wave rectifier output not contain first harmonic, which is more difficult to filter?

Your paper should be at least two pages, typed single-spaced, using 12-point font and 1-inch margins on the top, bottom, right, and left. You should use the APA guidelines for writing and citations.

Keep the following points in mind:

·         The efficiency of a buck DC-DC converter.

·         The principle of a buck DC-DC converter.