# Full Wave Rectifier Efficiency and Their Applications in Real-Life

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## What is Full Wave Rectifier Efficiency?

The efficiency of a full wave rectifier is the ratio of DC output power to the AC input power. The efficiency of full-wave rectifier is 81.2%. The rectification efficiency of full wave rectifiers is double that of half wave rectifiers. The ripple factor in full wave rectifiers is low hence a simple filter is required. The value of ripple factor in full wave rectifier is 0.482 while in half wave rectifier it is about 1.21.

The full wave rectifier circuit can be constructed in two ways, using a center-tapped transformer and two diodes or using a bridge rectifier. The efficiency of a full-wave rectifier is generally higher than that of a half-wave rectifier because it utilizes both halves of the AC waveform, resulting in a smoother DC output.

1. What is Full Wave Rectifier Efficiency?
2. Formula Used Efficiency of Full Wave Rectifier
3. How to Calculate of Full Wave Rectifier Efficiency?
4. Form Factor in Full Wave Rectifier Efficiency
5. Peak Factor in Full Wave Rectifier Efficiency
6. Applications of Full Wave Rectifier
• How does a full-wave rectifier differ from a half-wave rectifier?
• What is the maximum efficiency of full wave rectifier?
• What is minimum efficiency of full wave rectifier?
• How does the efficiency of a full-wave rectifier compare to a half-wave rectifier?
• What is the efficiency formula for a full-wave rectifier?
• What factors affect the efficiency of a full-wave rectifier?
• Can full-wave rectifier be used for high-frequency applications?
• How can the efficiency of a full-wave rectifier be improved?
• What is the full wave rectifier rectification efficiency derivation?

## Formula Used Efficiency of Full Wave Rectifier

The efficiency (η) of a full-wave rectifier can be calculated using the following formulas:

### DC Power Output (Pdc):

Pdc=Vdc×Idc

Where:

• Vdc is the average DC voltage across the load resistor.
• Idc is the DC current flowing through the load resistor.

Maximum AC Power Input (Pac):

Pac=Vmax × Imax

Where:

• Vmax is the peak AC voltage of the input waveform.
• Imax is the peak AC current flowing through the load resistor.

Efficiency (η):

η= Pdc/Pac×100%

You may need to consider factors such as diode losses, transformer losses, and other non-idealities in the circuit for a more accurate efficiency calculation.

## How to Calculate of Full Wave Rectifier Efficiency?

Let’s go through an example of calculating the efficiency of a full-wave rectifier. For simplicity, we’ll assume ideal conditions. Here are the given values:

• AC voltage (Vmax) = 12 V (peak AC voltage)
• Load resistance (R) = 1.5 kΩ (ohms)
• Diode voltage drop (Vd) = 0.7 V (for silicon diodes)
• Frequency (f) = 50 Hz

### Calculate the DC Voltage (Vdc):

The average DC voltage for a full-wave rectifier can be calculated as approximately 0.9 times the peak AC voltage.

Vdc ≈ 0.9 × Vmax

Vdc ≈ 0.9 × 12V ≈ 10.8V

Calculate the DC Current (Idc):

Using Ohm’s Law (I = V/R):

Idc = Vdc/R = 10.8V/1500Ω   ≈ 0.0072A = 7.2mA

Calculate DC Power Output (Pdc):

Pdc = Vdc × Idc ≈ 10.8V × 0.0072A ≈ 0.07776W ≈ 77.76mW

Calculate Maximum AC Power Input (Pac):

Since this is a full-wave rectifier, Pac is the same as the maximum AC power.

Pac=Vmax × Vmax/R

Pac=12V × 1500Ω12V ≈ 0.096W≈96mW

Calculate Efficiency (η):

η= Pdc/ Pac ×100%

η=77.76mW/96mW × 100% ≈ 80.75%

## Form Factor in Full Wave Rectifier Efficiency

The form factor (F.F.) of a full-wave rectifier is a measure of the shape of the rectified output waveform. The form factor is defined as the ratio of the root mean square (RMS) value of the output voltage to its average (DC) value.

For a full-wave rectified sinusoidal waveform, the RMS value (Vrms) is related to the peak value (Vmax) by the factor 0.9, and the average value (Vdc) is approximately 0.90.9 times Vmax in ideal conditions.

### The form factor is calculated using the following formula:

F.F. = RMS value/Average value

### For a full-wave rectified sinusoidal waveform:

F.F. = 0.9⋅Vmax/0.9.0.9⋅Vmax = 1/0.9 ≈ 1.11

So, the form factor for a full-wave rectifier is approximately 1.11. This value indicates that the RMS value of the output waveform is1.11 times its average (DC) value.

## Peak Factor in Full Wave Rectifier Efficiency

The peak factor (P.F.) of a full-wave rectifier is a measure of the ratio of the peak value of the output voltage to its root mean square (RMS) value. It provides information about the waveform’s peakiness or the presence of spikes in the signal.

For a full-wave rectified sinusoidal waveform, the RMS value (Vrms) is related to the peak value (Vmax) by the factor 0.9.

### The peak factor is calculated using the following formula:

P.F. = Peak value/RMS value

### For a full-wave rectified sinusoidal waveform:

P.F. = Vmax/0.9⋅Vmax = 0.91 ≈ 1.11

So, the peak factor for a full-wave rectifier is approximately 1.11. This value indicates that the peak value of the output waveform is 1.11 times its RMS value.

## Applications of Full Wave Rectifier in Daily Life

There are some common applications of full-wave rectifiers, which are going to use in different domains to fulfil several objectives, including:

Power Supplies: Full-wave rectifiers are commonly used in power supplies to convert AC voltage from the mains into a smoother DC voltage suitable for powering electronic devices.

Battery Charging: Full-wave rectifiers are employed in battery charging systems to convert AC from a power source (such as a wall outlet) into DC for charging batteries efficiently.

Audio Amplifiers: In audio amplifiers, full-wave rectifiers are used to convert the AC signal from microphones or other input sources into DC, which can be more easily amplified and processed.

Signal Demodulation: Full-wave rectifiers are used in communication systems for demodulating amplitude-modulated (AM) signals. The rectifier extracts the original audio or data signal from the modulated carrier wave.

Motor Control: In certain motor control applications, DC power is required for driving motors. Full-wave rectifiers can provide the necessary DC voltage for motor control circuits.

Light Dimmers: Full-wave rectifiers can be employed in light dimmer circuits where a variable DC voltage is needed to control the intensity of lights.

Welding Power Supplies: Full-wave rectifiers are used in welding machines to convert AC power from the mains into DC power for welding operations.

X-Ray Machines: X-ray machines often use full-wave rectifiers to convert AC power into high-voltage DC power for generating X-rays.

Uninterruptible Power Supplies (UPS): Full-wave rectifiers are employed in the rectification stage of UPS systems, ensuring a continuous and smooth DC power supply even in the event of a power outage.

Railway Traction Systems: Full-wave rectifiers are used in the power supply systems of electric trains and trams to convert AC power from overhead lines into DC power for driving the traction motors.

### How does a full-wave rectifier differ from a half-wave rectifier?

Half-wave rectifier only rectifies one-half (either positive or negative) of the AC waveform; whereas full-wave rectifier rectifies both halves. As a result, the full-wave rectifier provides the more constant and smoother DC output as compared to half-wave rectifier.

### What is the maximum efficiency of full wave rectifier?

The maximum efficiency of a full wave rectifier is 81.2%. This efficiency is achieved when the forward resistance of the diode is minimal.

### What is minimum efficiency of full wave rectifier?

The minimum efficiency of a full wave rectifier is zero. This occurs when the forward resistance of the diode is high, leading to a significant power loss and, consequently, zero efficiency.

### How does the efficiency of a full-wave rectifier compare to a half-wave rectifier?

The efficiency of full-wave rectifier is higher as compared to half-wave rectifier. This is because a full-wave rectifier implements both halves of the AC waveform, then resulting in a more continuous and less pulsating DC output.

### What is the efficiency formula for a full-wave rectifier?

The efficiency (η) of a full-wave rectifier can be calculated using the formula: η = (DC power output / AC power input) x 100%.

### What factors affect the efficiency of a full-wave rectifier?

The efficiency of full-wave rectifier is influenced by factors like as the forward voltage drop of diodes, transformer losses, and conduction losses in the diodes.

### Can full-wave rectifier be used for high-frequency applications?

Yes! Full-wave rectifiers are commonly used in high-frequency applications. However, the choice between center-tapped and bridge rectifiers may depend on specific design requirements.

### How can the efficiency of a full-wave rectifier be improved?

The efficiency of full-wave rectifier can be improved by using diodes with lower forward voltage drops, minimizing transformer losses, and optimizing the circuit design for decreased the conduction losses.

### What is the full wave rectifier rectification efficiency derivation?

Rectification efficiency (η) in a full-wave rectifier is the ratio of average DC power output to maximum AC power input. Derive it by integrating DC power and AC power expressions.

The rectification efficiency (η) formula for a full-wave rectifier is:

η = 0.9/1 + (4fRtCd)2

Where:

• f is the frequency of the AC input,
• Rt is the transformer secondary coil resistance,
• Cd is the filter capacitor value.

## Final Words

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