distinguish graphically between half-wave and full-wave rectification

Rectification and Smoothing – Cambridge IGCSE/A‑Level (9702) 21.1

Objective

To distinguish graphically between half‑wave and full‑wave rectification, to recognise the two practical full‑wave configurations (centre‑tap and bridge), and to understand why a smoothing capacitor works better with a full‑wave rectifier.

1. Introduction

• Rectification: conversion of an alternating‑current (AC) voltage into a unidirectional (pulsating) voltage.
• Two basic types:
  • Half‑wave rectifier – a single diode conducts only during one half‑cycle.
  • Full‑wave rectifier – both half‑cycles are used; the negative half‑cycle is inverted so the output always has the same polarity.
• The pulsating output can be smoothed with a capacitor to obtain a near‑DC supply.

2. Half‑Wave Rectifier

2.1 Circuit

_in R_L Ground

2.2 Waveforms (graphical)

_in = V_m sin ωt V_out (half‑wave) 0 T

2.3 Key Features

• Output exists for one half of each input period → ripple frequency = f (same as mains).
• Peak inverse voltage (PIV) on the diode = V_m.
• Ideal conversion efficiency ≈ 40 % (ratio of DC power delivered to AC power supplied).
• Only one diode is required; no centre‑tap transformer needed.

3. Full‑Wave Rectifier

3.1 Practical Configurations

3.1.1 Centre‑Tap Transformer (2‑diode)

_L Ground

3.1.2 Bridge (Full‑Bridge) Rectifier (4‑diode)

_in R_L Ground

3.2 Waveforms (graphical)

_in V_out (full‑wave) 0 T

3.3 Key Features

• Output present for both halves of the input → ripple frequency = 2f (twice the mains frequency).
• Peak inverse voltage (PIV):
  • Bridge: PIV = V_m (each diode sees the peak of the input).
  • Centre‑tap: PIV = 2 V_m (each diode must block the full secondary voltage).
• Ideal conversion efficiency ≈ 81 % (significantly higher than half‑wave).
• Requires either a centre‑tapped transformer (2 diodes) or a bridge network (4 diodes).

3.4 Comparative Table

Feature	Half‑Wave	Full‑Wave (Centre‑Tap)	Full‑Wave (Bridge)
Diodes required	1	2	4
Transformer	Single‑ended (no centre tap)	Centre‑tap secondary	Standard (non‑centre‑tapped) secondary
Output during each input cycle	1 positive pulse	2 positive pulses	2 positive pulses
Ripple frequency	f (same as mains)	2f	2f
Peak Inverse Voltage (PIV)	Vm	2 Vm	Vm
Ideal conversion efficiency	≈ 40 %	≈ 81 %	≈ 81 %
Typical applications	Low‑power signalling, simple chargers	Medium‑power supplies where a centre‑tap transformer is already available	Most modern DC supplies, especially where transformer size/weight is a concern

4. Smoothing with a Capacitor

4.1 Why Full‑Wave Is Preferred

The smoothing capacitor is charged to the peak of each pulse and then discharges through the load between pulses. The discharge interval is the reciprocal of the ripple frequency. Because a full‑wave rectifier doubles the ripple frequency, the capacitor has less time to discharge, producing a smaller ripple voltage for the same capacitance.

4.2 Ripple‑Voltage Formula

Assuming a resistive load that draws a constant current I:

\[ V_{\rm r}= \frac{I}{f\,C} \]

• f = ripple frequency (half‑wave: f, full‑wave: 2f)
• C = capacitance (farads)
• I = load current (amperes)

Consequently, for a full‑wave rectifier the ripple voltage is halved:

\[ V_{\rm r(full)} = \frac{I}{2f\,C}= \frac{V_{\rm r(half)}}{2} \]

4.3 Ripple Factor

The ripple factor (k) expresses the relative size of the residual AC component:

\[ k = \frac{V_{\rm r(pp)}}{V_{\rm dc}} \]

• For an ideal half‑wave rectifier with a large filter capacitor, k ≈ 0.48 / √(f C) (approximation).
• For an ideal full‑wave rectifier, k ≈ 0.31 / √(2f C), i.e. roughly half the half‑wave value.

4.4 Worked Example

Given:

• Mains: 240 V r.m.s., 50 Hz.
• Full‑wave bridge rectifier.
• Load current: I = 100 mA.
• Filter capacitor: C = 100 µF.

Step 1 – Peak secondary voltage (ignoring transformer losses):

\[ V_{m}= \sqrt{2}\times 240\;\text{V} \approx 340\;\text{V} \]

Step 2 – Ripple frequency:

\[ f_{\rm ripple}=2f = 2\times 50\;\text{Hz}=100\;\text{Hz} \]

Step 3 – Ripple voltage (using the formula for a full‑wave rectifier):

\[ V_{\rm r}= \frac{I}{2f\,C}= \frac{0.10\;\text{A}}{100\;\text{Hz}\times 100\times10^{-6}\;\text{F}} = \frac{0.10}{0.01}=10\;\text{V (peak‑to‑peak)} \]

Step 4 – Approximate DC output (ignoring diode drops):

\[ V_{\rm dc}\approx V_{m}-\frac{V_{\rm r}}{2}=340\;\text{V}-5\;\text{V}=335\;\text{V} \]

Result: With a 100 µF capacitor the ripple is only 10 V p‑p, giving a relatively smooth DC output of about 335 V.

5. Summary

• Half‑wave rectifiers use a single diode; only one half‑cycle contributes to the output, giving a ripple frequency equal to the mains frequency.
• Full‑wave rectifiers (centre‑tap or bridge) use two or four diodes to utilise both half‑cycles, doubling the ripple frequency and halving the ripple voltage for a given capacitor.
• The higher ripple frequency of full‑wave rectifiers makes the smoothing capacitor more effective, which is why full‑wave designs are preferred in most DC‑power supplies.
• Key design parameters to compare are diode count, transformer requirement, PIV, ripple frequency, conversion efficiency, and typical applications (see the comparative table).