Physics › Magnetic fields › Rectification and smoothing
Rectification and smoothing
The grid delivers ac, and almost everything you own wants dc. The conversion happens inside every charger you plug in: diodes force the current one way, and a capacitor fills in the gaps. CIE examines it directly; for everyone else it is the electronics your phone survives on.
Builds on Alternating currents and Charging and discharging.
IN THIS TOPIC
- Sketch the output of half-wave and full-wave rectification of a sinusoidal input.
- Explain how four diodes in a bridge steer both half-cycles the same way through a load.
- Explain smoothing with a reservoir capacitor, and how C and the load resistance set the ripple.
WHAT YOU PROBABLY THINK
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, and reversed it passes almost nothing. Put a single diode in series with an ac supply and a load, and only the half-cycles that point 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, which is why the 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, so nothing is wasted, 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: the negative half-cycles are folded up rather than discarded, giving twice as many humps and no dead time. It is one-way, but it is nothing like steady, and a circuit that wants dc wants the bumps gone 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: 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: 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 from 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: the reservoir must be sized for the heaviest load the circuit will ever draw.
THE EXAM BIT
- Sketches score for the right shape: half-wave keeps gaps at zero, full-wave folds the negative halves up with no gaps, and the smoothed trace is a sawtooth-topped near-flat line hugging the peaks.
- In a bridge question, identify which two diodes conduct in each half-cycle and mark the current direction through the load: it must be the same both times, or the bridge has failed on paper.
- Explain smoothing in the capacitor's own vocabulary: charges near the peaks, discharges through the load between them, with the time constant RC against the time between peaks deciding the ripple.
- Larger C or larger load resistance means smaller ripple. State the comparison rather than the bare direction: RC long compared with the interval between peaks.
- This topic is examined by CIE; 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 it is much less steady: 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.
CHECK YOUR PROGRESS
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- Sketch the output of half-wave and full-wave rectification of a sinusoidal input.
- Explain how four diodes in a bridge steer both half-cycles the same way through a load.
- Explain smoothing with a reservoir capacitor, and how C and the load resistance set the ripple.
Open the full revision checklist to track your progress across the whole unit.
No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.