All Exams Test series for 1 year @ ₹349 only
Question

The maximum rectification efficiency in case of a full wave rectifier is

The correct answer is

81.2%

Rectification Efficiency Explained

Rectification efficiency is a key measure used in electronics to determine how effectively an AC voltage is converted into a DC voltage. It is defined as the ratio of the DC power output ($P_{dc}$) delivered to the load resistor ($R_L$) to the AC power input ($P_{ac}$) drawn from the source (usually a transformer secondary).

The formula for rectification efficiency ($\eta$) is:

$$ \eta = \frac{P_{dc}}{P_{ac}} \times 100\% $$

Full-Wave Rectifier Basics

A full-wave rectifier is designed to utilize both the positive and negative half-cycles of the AC input waveform. This process results in a pulsating DC output with a lower ripple factor and a higher average DC value compared to a half-wave rectifier, making it more efficient in converting AC to DC.

Calculating Rectification Efficiency

The efficiency of a full-wave rectifier depends on the load resistance ($R_L$) and the internal forward resistance ($r_f$) of the diodes used. The general formula derived from power calculations is:

$$ \eta = \frac{I_{dc}^2 R_L}{I_{rms}^2 (R_L + r_f)} \times 100\% $$

For a full-wave rectifier with a resistive load, the DC component of the current is $I_{dc} = \frac{2I_m}{\pi}$ and the RMS value of the current is $I_{rms} = \frac{I_m}{\sqrt{2}}$, where $I_m$ is the peak current through the load.

Substituting these into the efficiency formula:

$$ \eta = \frac{(\frac{2I_m}{\pi})^2 R_L}{(\frac{I_m}{\sqrt{2}})^2 (R_L + r_f)} \times 100\% $$

Simplifying this expression gives:

$$ \eta = \frac{\frac{4I_m^2}{\pi^2} R_L}{\frac{I_m^2}{2} (R_L + r_f)} \times 100\% $$

$$ \eta = \frac{8}{\pi^2} \frac{R_L}{R_L + r_f} \times 100\% $$

Maximum Efficiency Derivation

The maximum possible rectification efficiency for a full-wave rectifier occurs under ideal conditions. These ideal conditions assume that the internal forward resistance of the diodes ($r_f$) is zero ($r_f = 0$), meaning the diodes offer no resistance to current flow when forward-biased.

Under these ideal conditions ($r_f = 0$), the efficiency formula becomes:

$$ \eta_{max} = \frac{8}{\pi^2} \times 100\% $$

Now, we calculate the numerical value:

Using the approximate value of $\pi \approx 3.14159$, we get $\pi^2 \approx 9.8696$.

$$ \eta_{max} \approx \frac{8}{9.8696} \times 100\% $$

$$ \eta_{max} \approx 0.81197 \times 100\% $$

$$ \eta_{max} \approx 81.2\% $$

This value represents the theoretical upper limit for the rectification efficiency of a full-wave rectifier circuit.

Conclusion

Based on the derivation for ideal conditions, the maximum rectification efficiency achievable by a full-wave rectifier is approximately 81.2%.

Was this answer helpful?

Important Questions from Rectifier Circuits

  1. What is the ripple factor of full-wave bridge rectifier?

  2. The maximum efficiency of a half-wave rectifier is

  3. For a full wave rectifier, the output frequency

  4. A full wave rectifier is supplied from a $20$ V AC supply. Average output voltage is:
  5. A half wave rectifier requires -

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App