Physics › Magnetic fields › Rectification and smoothing
Rectification and smoothing
The mains supply is alternating and almost every device runs on direct current, so the conversion happens inside every charger. Diodes force the current one way, half-wave with one and full-wave with four in a bridge, and a capacitor across the output fills in the gaps between peaks. This is examined directly by CIE.
Pick your board and the few notes written for the other boards quietly fold away, here and in the practice players. Nothing is deleted: every folded piece reopens on a tap.
Builds on Alternating currents and Charging and discharging.
IN THIS TOPIC
- Sketch half-wave and full-wave rectified outputs, and explain how a bridge of four diodes steers both half-cycles the same way through a load.
- Explain smoothing with a reservoir capacitor, and how C and the load resistance together set the ripple.
COMMON MISCONCEPTION
Rectified ac is direct current, so it is steady.
One-way traffic
A diode conducts in one direction only. A small forward voltage switches it on; reversed, it passes almost nothing. Put a single diode in series with an ac supply and a load and only the half-cycles pointing the diode's way get through. The output is half-wave rectified, humps of one polarity with a flat nothing where every other half-cycle used to be.
Half the supply is thrown away, so a single diode is mostly a teaching circuit. The fix is to steer the negative half-cycles through the load in the same direction as the positive ones, wasting nothing, and that takes four diodes arranged as a bridge.
The bridge rectifier
In a bridge, the four diodes work in opposite pairs. During one half-cycle, two of them conduct and route the current down through the load; during the other half-cycle the other two conduct, and the current from the reversed supply is routed down through the load the same way. The load never finds out that the supply changed sign.
The output is full-wave rectified, with the negative half-cycles folded up instead of discarded, giving twice as many humps and no dead time. It runs one way, but it is nothing like steady, and a circuit that needs dc needs the ripple removed too.
Smoothing
The cure is a reservoir capacitor across the load. Near each peak the supply tops the capacitor up; as the rectified voltage falls away, the capacitor discharges through the load and holds the voltage up until the next hump arrives to recharge it. The output becomes a slightly wobbly dc, and the wobble has a name of its own, ripple.
How much ripple survives is a time-constant argument from the capacitance unit. The capacitor discharges with time constant RC, where R is the load. Make RC long compared with the time between peaks and the capacitor barely sags before rescue arrives; the ripple is small. A larger capacitance smooths better, and so does a lighter load, meaning a larger load resistance drawing less current.
WORKED EXAMPLE
Is the smoothing any good?
A full-wave rectified 50 Hz supply is smoothed by a 470 μF capacitor across a 1.0 kΩ load. Compare the discharge time constant with the time between peaks.
Full-wave rectification doubles the humps, so peaks arrive every half-cycle, one every 10 ms.
RC = 1000 × 470 × 10−6 = 0.47 s, which is 47 times the 10 ms gap.
The capacitor loses only a small fraction of its charge before the next top-up, so the ripple is a few per cent of the peak. Respectable dc, out of an ac wall socket.
Drop the load resistance and the same capacitor sags further between peaks, so the ripple grows. That is the design trade at the heart of every power supply. Size the reservoir for the heaviest load the circuit will ever draw.
ASSESSMENT FOCUS
- Sketches score for the right shape. Half-wave keeps gaps at zero, full-wave folds the negative halves up with no gaps at all, and the smoothed trace is a near-flat line with a sawtooth top that hugs the peaks.
- In a bridge question, identify which two diodes conduct in each half-cycle and mark the current direction through the load. It has to come out the same both times, or your bridge has failed on paper.
- Explain smoothing in the capacitor's own vocabulary. It charges near the peaks and discharges through the load between them, with RC measured against the time between peaks deciding the ripple. Larger C or larger load resistance means smaller ripple, and stating that comparison beats stating the bare direction.
- CIE examines this topic directly, while AQA, Edexcel and OCR A treat it as background. Know your own specification before you spend revision time here.
CHECK YOURSELF
A student replaces the 470 μF reservoir capacitor in a full-wave supply with a 47 μF one. State and explain what happens to the output.
Show a hint
Think in time constants against the 10 ms between peaks.
Show the answer
The ripple grows by a factor of about ten. RC falls from 0.47 s to 0.047 s, so between top-ups the capacitor discharges ten times as fast through the same load and its voltage sags ten times further.
The output is still dc in direction, but far less steady, and the trace dips visibly between peaks instead of hugging them.
Diodes make the current one-way; the bridge wastes neither half.
The reservoir capacitor rides over the gaps; ripple shrinks as RC grows.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their mark schemes on the rectification and smoothing questions page.
CHECK YOUR PROGRESS
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- Sketch half-wave and full-wave rectified outputs, and explain how a bridge of four diodes steers both half-cycles the same way through a load.
- Explain smoothing with a reservoir capacitor, and how C and the load resistance together set the ripple.
Open the full revision checklist to track your progress across the whole unit.