explain the use of four diodes (bridge rectifier) for the full-wave rectification of an alternating current

Rectification and Smoothing

Objective

Explain how four diodes are used in a bridge configuration to achieve full‑wave rectification of an alternating current (AC) source.

1. Why Full‑Wave Rectification?

• Converts both the positive and negative halves of the AC waveform into a unidirectional (positive) output.
• Reduces the ripple frequency to twice the mains frequency, making filtering easier.
• Provides higher average output voltage compared with half‑wave rectification.

2. Bridge Rectifier Circuit

The bridge rectifier uses four diodes (D1–D4) arranged in a diamond shape. The AC source is connected across the two opposite corners, and the load is connected across the remaining two corners.

3. Operation During One AC Cycle

1. Positive half‑cycle (\$v_{in}>0\$)

   • D1 and D2 become forward‑biased and conduct.
   • D3 and D4 are reverse‑biased and block.
   • Current flows from the AC source through D1, the load (positive terminal), D2, and back to the source.
   • The output voltage across the load is \$V{out}=V{in}-2V{D}\$, where \$V{D}\$ is the forward voltage drop of a diode.

2. Negative half‑cycle (\$v_{in}<0\$)

   • D3 and D4 become forward‑biased, D1 and D2 reverse‑biased.
   • Current now flows through D3, the load (still with the same polarity), D4, and back to the source.
   • The magnitude of \$V{out}\$ remains the same as in the positive half‑cycle (again reduced by \$2V{D}\$).

4. Output Waveform

The resulting output is a series of positive half‑sine pulses. The frequency of the ripple is \$2f\$, where \$f\$ is the mains frequency (e.g., \$2\times50\ \text{Hz}=100\ \text{Hz}\$).

5. Ripple \cdot oltage and Smoothing

To obtain a near‑DC voltage, a filter capacitor \$C\$ is placed across the load. The ripple voltage \$V_{ripple}\$ for a resistive load \$R\$ can be approximated by

\$V{ripple}\approx\frac{I{load}}{2fC}=\frac{V_{DC}}{R\;2fC}\$

where \$I{load}=V{DC}/R\$ and \$V{DC}\approx V{peak}-2V_{D}\$.

6. Example Calculation

From the data:

• \$V{DC}\approx V{peak}-2V_{D}=15-1.4=13.6\ \text{V}\$
• \$I{load}=V{DC}/R=13.6\ \text{V}/1\ \text{kΩ}=13.6\ \text{mA}\$
• \$V{ripple}\approx\dfrac{I{load}}{2fC}= \dfrac{13.6\times10^{-3}}{2\times50\times1000\times10^{-6}} \approx 0.136\ \text{V}\$

The ripple is only about 0.14 V, giving a smooth DC output suitable for most low‑power electronics.

7. Advantages of the Bridge Rectifier

• Uses both halves of the AC cycle → higher efficiency.
• No need for a centre‑tapped transformer, reducing cost and size.
• Higher ripple frequency simplifies filtering.

8. Summary

A bridge rectifier employing four diodes converts an AC input into a full‑wave rectified output. The resulting pulsating DC can be smoothed with a capacitor, where the ripple voltage is inversely proportional to the product of ripple frequency and capacitance. Understanding the operation and calculation of ripple is essential for designing reliable power supplies in A‑Level physics experiments.