A2 · Practice questions · Rectification and smoothing

One way through the diode.

Six original Cambridge-style questions on half-wave and full-wave rectification, the bridge of four diodes, ripple frequency, and choosing a smoothing capacitor from the time constant RC.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Four diodes, then a capacitor across the load.

Rectification: one diode passes only the half-cycles that forward-bias it (half-wave); a bridge of four routes both halves the same way through the load (full-wave).
Ripple: half-wave output ripples at the supply frequency, full-wave at twice it; the peak output is V₀ = √2 V_rms for ideal diodes.
Smoothing: a capacitor in parallel with the load charges to the peak and discharges through R between peaks; a larger RC compared with the ripple period means a smaller ripple.

Analysis

[3 marks]

Explain how a bridge of four diodes connected to a sinusoidal supply produces full-wave rectification of the current in a load resistor. State the ripple frequency of the output when the supply frequency is 60 Hz.

During the half-cycle when the supply terminal A is positive, the two diodes that are forward biased conduct and steer the current through the load in one direction; the other two are reverse biased and block. ✓
During the other half-cycle the roles swap: the other pair conducts, and the current is again steered through the load in the same direction, so every half-cycle appears across the load as a positive hump and the output is zero only at instants. ✓
There are two humps per supply cycle, so the ripple frequency is 2 × 60 = 120 Hz. ✓

Calculation

[2 marks]

A transformer secondary supplies 15 V r.m.s. at 50 Hz to a rectifier with ideal diodes. Calculate the peak output p.d. across the load, and state the time between successive peaks of the output for half-wave and for full-wave rectification.

V₀ = √2 × V_rms = 1.414 × 15 = 21 V. ✓
Half-wave: one peak per cycle, so 1/50 = 20 ms; full-wave: two per cycle, so 10 ms. ✓

Calculation

[3 marks]

A full-wave rectifier fed from a 50 Hz supply gives a peak output of 12 V. A 470 μF smoothing capacitor is connected across a 1.0 kΩ load. Calculate the time constant of the capacitor and load, and estimate the ripple (the fall in p.d. between one peak and the next), taking the capacitor to discharge through the load for the whole interval between peaks.

τ = RC = 1000 × 470 × 10⁻⁶ = 0.47 s. ✓
Between full-wave peaks is 10 ms, so the p.d. falls to 12 × e^{−0.010/0.47} = 12 × 0.979 = 11.75 V. ✓
Ripple = 12 − 11.75 = 0.25 V, about 2% of the peak, because the time constant is much longer than the 10 ms between peaks. ✓

Analysis

[3 marks]

A student connects a smoothing capacitor in series with the load resistor at the output of a rectifier, and finds that no steady current flows in the load. Explain why, and state the correct arrangement. Explain also why, with the capacitor correctly connected, the ripple becomes larger when the load draws a larger current.

In series, the capacitor charges up and then blocks any steady current, because no charge can pass through its dielectric; it only passes the changing part of the p.d. ✓
The capacitor must be connected in parallel with (across) the load, so that it charges to the peak and then supplies the load from its stored charge between peaks. ✓
A larger load current means a smaller load resistance and so a shorter time constant RC; the capacitor loses more of its charge between peaks, so the p.d. dips further and the ripple is larger. ✓

Calculation

[3 marks]

A half-wave rectifier fed from a 50 Hz supply has a 2.2 kΩ load. A smoothing capacitor is to be chosen so that the output p.d. falls by no more than 5.0% between successive peaks. Taking the capacitor to discharge for the whole time between peaks, calculate the minimum time constant needed and hence the minimum capacitance.

Half-wave peaks are 20 ms apart, and the p.d. must stay above 0.95 of the peak: e^−0.020/RC = 0.95, so 0.020 / RC = −ln 0.95 = 0.0513. ✓
RC = 0.020 / 0.0513 = 0.39 s. ✓
C = 0.39 / 2200 = 1.8 × 10⁻⁴ F, so a 180 μF capacitor (or the next standard value above, 220 μF) is needed. ✓

Calculation

[4 marks]

A low-voltage power supply uses a transformer giving 6.0 V r.m.s. at 50 Hz, a bridge rectifier with ideal diodes, and a 1000 μF smoothing capacitor across a 100 Ω load.

(a) Calculate the peak p.d. across the load. (b) State the ripple frequency and the time between peaks. (c) Calculate the time constant of the capacitor and load, and estimate the lowest p.d. the output falls to between peaks and the size of the ripple.

(a) V₀ = √2 × 6.0 = 8.5 V. ✓
(b) Full-wave rectification doubles the frequency: ripple at 100 Hz, so peaks are 10 ms apart. ✓
(c) τ = RC = 100 × 1000 × 10⁻⁶ = 0.10 s; the p.d. falls to 8.49 × e^{−0.010/0.10} = 8.49 × 0.905 = 7.7 V. ✓
Ripple = 8.49 − 7.68 = 0.8 V; a larger capacitor or a smaller load current would reduce it. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.