Salam Academy
Sign inCreate account
Learn7408Paper 3B · Option: ElectronicsDiscrete semiconductor devicesRectification and smoothing
Discrete semiconductor devices · Subtopic 1 of 1

Rectification and Smoothing

Rectification converts a.c. to d.c. using diodes, which conduct in one direction only. A single diode gives half-wave rectification: only alternate half-cycles pass. A bridge of four diodes gives full-wave rectification: both halves pass, in the same direction through the load. A capacitor in parallel with the load smooths the output: it charges near each peak and discharges slowly through the load in between. Larger RCRC (compared with the time between peaks) means less ripple.

Cambridge 9702 Paper 4AQA 7408 Paper 3B≈ 35 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Sketch and distinguish half-wave and full-wave rectified outputs.
  • Explain how a single diode and a bridge rectifier work.
  • Explain smoothing by a capacitor.
  • Describe how CC and the load resistance affect the ripple.
Pause and think

Your phone charger plugs into 230230 V a.c. but your phone needs about 55 V d.c. What has to happen inside the charger?

Section 2 of 14

Core idea

Start hereCharging a phone from the wall

The wall socket provides a.c., but your phone's battery needs d.c. Inside the charger, diodes chop and flip the waveform so current only flows one way, and a capacitor smooths out the bumps.

The big idea

Half-wave: one diode, half of each cycle. Full-wave bridge: four diodes flip negative halves. Smoothing capacitor: RC≫RC\gg time between peaks.

1. Rectification

Half-wave, full-wave and smoothed output
24681012−0.20.20.40.60.811.20timeoutput Vfull-wavehalf-wavesmoothed
A larger RCRC (bigger capacitor or load resistance) makes the capacitor discharge more slowly, so the ripple is smaller.
Bridge rectifier current paths

Two diodes conduct at a time; the load always sees the same polarity.

2. Smoothing

Ripple depends on RC
  1. V=V0e−t/RCV=V_0e^{-t/RC}
    Between peaks, the capacitor discharges through the load.
  2. t≪RC⇒V≈V0t\ll RC\Rightarrow V\approx V_0
    Little decay before the next peak recharges it.
Worked exampleChoosing C

Full-wave 50 Hz, load 1 kΩ.

Goal
Suitable CC.
Strategy
RC≫10RC\gg10 ms.
  1. peaks every 10 ms\text{peaks every }10\text{ ms}
    Full-wave doubles the frequency to 100 Hz.
  2. C=100 μF⇒RC=0.1 sC=100\ \mu\text{F}\Rightarrow RC=0.1\text{ s}
    10× longer: ripple about e−0.1e^{-0.1}, ~10%.
Answer100100 μF (bigger for less ripple).
✓ CheckTrade-off: larger, costlier, big charging surges ✓.
Common misconceptionRipple frequency Full-wave rectified 50 Hz ripples at 100 Hz.

Summary

  • Half vs full wave.
  • Smoothing: large RCRC.
Key words
Ripple
Remaining variation after smoothing.
HistoryCrystal sets. Early radio listeners in the 19101910s used "cat's whisker" detectors: a thin wire touching a galena crystal formed a crude diode that rectified radio signals. No battery was needed; the radio wave's own energy drove the earphone.
Fun factThe 100 Hz hum. If a cheap power supply's smoothing capacitor ages and loses capacitance, the ripple increases and you can hear a faint 100100 Hz hum through speakers. It's the full-wave rectified mains frequency leaking through.
Keep reading for free

Create a free account to read the rest of this lesson: key formulae, worked examples, explore the graph, common mistakes, exam tip, guided practice and more. Your answers are checked and your progress is saved.

Create free accountSign in