Physics › Magnetic fields › Rectification and smoothing
Rectification and smoothing
Diodes conduct one way only, so one diode gives half-wave rectification and a bridge of four gives full-wave, using both halves of the cycle. A reservoir capacitor across the load charges at each peak and discharges between them, and the ripple falls as the time constant RC grows.
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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.
Rectified current is one-directional but not steady: it still rises and falls each half-cycle. A reservoir capacitor smooths it towards steady dc, and some ripple always remains.
Half-wave rectification
A diode conducts in one direction only: a small forward voltage switches it on, and in reverse it passes almost no current. A single diode in series with an ac supply and a load passes only the half-cycles in the diode's forward direction. The output is half-wave rectified: peaks of one polarity, with zero output where the other half-cycles were.
Half of each cycle is wasted, so a single diode appears mainly in demonstrations. The alternative is to direct the negative half-cycles through the load in the same direction as the positive ones, which requires four diodes arranged as a bridge.
The bridge rectifier
In a bridge, the four diodes conduct in opposite pairs. During one half-cycle two of them conduct and route the current downwards through the load; during the other half-cycle the other two conduct, and the current from the reversed supply is routed downwards through the load as well. The current through the load is therefore in the same direction throughout.
The output is full-wave rectified, with the negative half-cycles inverted rather than discarded, giving twice as many peaks and no interval of zero output. The current is unidirectional but not steady, so a circuit requiring dc also needs the variation removed.
Smoothing
A reservoir capacitor across the load reduces that variation. Near each peak the supply charges the capacitor; as the rectified voltage falls, the capacitor discharges through the load and maintains the output voltage until the next peak recharges it. The output is then approximately steady dc, and the residual variation is called ripple.
The size of the ripple follows from the time constant met in the capacitance unit. The capacitor discharges with time constant RC, where R is the load resistance. If RC is long compared with the time between peaks, the capacitor voltage falls only slightly before the next peak and the ripple is small. A larger capacitance reduces the ripple, and so does a larger load resistance, which draws less current.
WORKED EXAMPLE
Comparing RC with the time between peaks
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 number of peaks, 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 it is recharged, so the ripple is a few per cent of the peak value.
Reducing the load resistance increases the current drawn, so the same capacitor discharges further between peaks and the ripple increases. A power supply's reservoir capacitor is therefore sized for the largest current the circuit will draw.
ASSESSMENT FOCUS
- Sketches are marked on shape: half-wave keeps intervals at zero, full-wave inverts the negative half-cycles with no intervals, and the smoothed trace is a near-flat line with a sawtooth variation following 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 in both half-cycles.
- Explain smoothing in terms of the capacitor: it charges near the peaks and discharges through the load between them, and the ripple depends on RC compared with the time between peaks. State that a larger C or a larger load resistance gives a smaller ripple, and compare RC with the time between peaks rather than giving only the direction of change.
- CIE examines this topic directly; AQA, Edexcel and OCR A do not. Check your own specification before revising it.
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
Compare the time constant RC with 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 current is still unidirectional, but the output is much less steady, and the trace falls visibly between peaks instead of following them closely.
A diode passes current in one direction only, and a bridge of four uses both half-cycles.
A reservoir capacitor discharges through the load between peaks; the ripple falls as RC increases.
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.
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