Analyse capacitor smoothing in a rectifier
Explain how a capacitor across the load smooths rectified output and how capacitance and load resistance affect ripple.
Section 21 · Alternating currents
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 21. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use period, frequency, sine graphs, current, voltage, power, diodes, capacitors and SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does the smoothing capacitor do between rectified peaks?
What you will learn
- Explain capacitor charging and discharging in a rectifier
- Analyse how capacitance changes ripple
- Analyse how load resistance changes discharge and ripple
The capacitor fills the gaps between rectified peaks
Connect a capacitor in parallel with the load resistor after the rectifier. Near each output peak, the rectifier voltage exceeds the capacitor voltage and the capacitor charges. As the rectified voltage falls below the capacitor voltage, the diode path stops conducting and the capacitor discharges through the load.
The discharge supplies the load between peaks, so the output voltage falls gradually rather than dropping to zero. The result is a more nearly steady voltage with a residual ripple. A capacitor smooths the waveform; it does not make the output perfectly constant.
Capacitance and load resistance set the ripple
For the same charge removed between peaks, a larger capacitance gives a smaller voltage change because ΔQ = CΔV. The capacitor therefore loses less voltage between peaks and ripple decreases when C increases.
A larger load resistance draws less current at a given voltage, so the capacitor discharges more slowly between peaks and ripple decreases. A smaller resistance draws more current and increases ripple. Full-wave rectification has more frequent charging peaks than half-wave rectification, which also leaves less time for discharge.
Pause and explain
If capacitance is increased with the same load and rectifier, what happens to ripple?
Worked example
A rectifier's smoothing capacitor is increased while the load stays the same. Predict and explain the change in output ripple.
Show the worked solution
- The load removes charge between peaks.
- For fixed charge removed, ΔV = ΔQ/C.
- Larger C means smaller voltage fall between peaks.
Answer Ripple decreases because the larger capacitor loses less voltage while supplying the load between charging peaks.
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Explain the capacitor's state near a rectified peak and during the following gap between peaks.
Give me a hint
Track when the source voltage is above or below the capacitor voltage.
Compare my reasoning
- Near a peak, the rectifier conducts and charges the capacitor.
- After the peak, the source voltage falls below the capacitor voltage and diode conduction stops.
- The capacitor then discharges through the load, keeping its voltage from falling as quickly as it otherwise would.
It charges near each peak and discharges through the load between peaks to support the output voltage.
Look for these in your work
- Charging interval identified
- Diode condition described
- Discharge effect explained
A smoothing circuit is changed from a 1.0 kΩ load to a 4.0 kΩ load, with the same capacitor and rectified input. Predict the ripple change and explain it.
Give me a hint
Use load current at a given voltage and the capacitor discharge interval.
Compare my reasoning
- The larger resistance draws less load current at a given voltage.
- Less charge leaves the capacitor between peaks in the same time.
- Its voltage falls less, so ripple decreases.
Ripple decreases because the larger load resistance draws less current and the capacitor discharges more slowly.
Look for these in your work
- Current-resistance relation used
- Charge loss compared
- Ripple direction stated
A student says a very large smoothing capacitor makes rectified output perfectly constant. Correct the claim and compare what happens if the load resistance is reduced.
Give me a hint
Distinguish smaller ripple from zero ripple and track discharge current.
Compare my reasoning
- A larger capacitor reduces the voltage drop between peaks but does not guarantee zero ripple.
- Reducing load resistance increases current at a given output voltage.
- The capacitor discharges faster between peaks, increasing ripple.
A larger capacitor makes the output steadier but residual ripple can remain; reducing load resistance increases ripple by drawing more current.
Look for these in your work
- Perfect-constant claim corrected
- Load-current change explained
- Ripple direction justified
Common mistakes
- Connecting the smoothing capacitor in series with the load
- Claiming a capacitor eliminates all ripple
- Saying a larger load resistance makes discharge faster
- Confusing capacitance with the load resistance
What can you explain now?
What does the smoothing capacitor do between rectified peaks?
Compare with the explanation
Discharges through the load and supplies it
Stored charge supports the load while input voltage is below capacitor voltage.
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Make it stick
How do capacitor size and load resistance affect voltage loss between rectified peaks?
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