Diode as a rectifier: half-wave and full-wave
A rectifier changes AC into DC using a diode's one-way conduction. Half-wave uses one diode; full-wave uses two diodes with a centre-tap transformer and doubles the output frequency.
Half-wave rectifier: how it works
- 1AC from the secondary of a transformer is applied to a diode in series with a load resistor RL.
- 2Positive half cycle: the diode is forward biased and conducts, so a voltage appears across RL.
- 3Negative half cycle: the diode is reverse biased and does not conduct, so the output is zero.
- 4The output is a row of positive half cycles with gaps: a pulsating DC with the same frequency as the input.
- 5Only half of each input cycle is used. The diode's breakdown voltage must be well above the peak AC voltage, or it will fail in the negative half cycle.
Full-wave rectifier: how it works
- 1A centre-tap transformer gives two equal voltages, from the centre tap C to each end A and B, which are opposite in phase.
- 2Diode D1 is joined to end A and diode D2 to end B. Their other ends are joined, and the load RL is connected between this joint and the centre tap C.
- 3First half cycle: A is positive with respect to C. D1 is forward biased and conducts; D2 is reverse biased.
- 4Next half cycle: B is positive with respect to C. D2 conducts; D1 is reverse biased.
- 5In both halves the current flows through RL in the same direction, so the output is positive in every half cycle.
- 6★There is one output pulse per half cycle, so the output frequency is twice the input: 100 Hz for 50 Hz mains.
★ key step
How to draw the circuits and waveforms
- Half-wave circuit: a transformer (two coils with lines for the core), one diode D and a load resistor RL in series with the secondary.
- Half-wave waveforms: an input sine wave, and below it an output with only the positive halves and flat gaps between them.
- Full-wave circuit: draw a transformer whose secondary has ends A and B and a centre tap C. Join D1 to A and D2 to B, and join their other ends at one point. Connect RL from that point back to C, and mark the current direction through RL.
- Full-wave waveforms: the input sine wave; the output from D1 and from D2 in alternate halves; and the combined output, a row of positive humps with no gaps.
- Keep the time axes of input and output lined up one below the other.
Smoothing the output with a capacitor
Rectified output is one-way but bumpy; the bumps are called ripple. A capacitor joined across RL smooths it. When the voltage rises, the capacitor charges up to the peak. When the voltage falls, it discharges slowly through RL and keeps the output up. So the output becomes nearly steady DC. A larger capacitor or a larger load resistance gives a smoother output.
Half-wave and full-wave rectifiers
| Feature | Half-wave | Full-wave |
|---|---|---|
| Diodes used | One | Two, with a centre-tap transformer |
| Part of input used | One half of each cycle | Both halves |
| Output frequency (50 Hz input) | 50 Hz | 100 Hz |
| Output shape | Pulses with gaps | Pulses with no gaps, smoother |
| Efficiency | Lower | Higher |
Worked example: output frequency and peak voltage
A 50 Hz supply is rectified. (a) What is the output frequency of a half-wave and of a full-wave rectifier? (b) In the full-wave rectifier, each half of the secondary gives 10 V rms. Find the peak output voltage, ignoring the diode drop, and then allowing 0.7 V for the diode.
- (a) Half-wave: one pulse per cycle, so 50 Hz. Full-wave: one pulse per half cycle, so 2 × 50 = 100 Hz.
- (b) Peak voltage = √2 × rms = 1.414 × 10 V ≈ 14.1 V.
- With a 0.7 V drop across the conducting diode: 14.1 − 0.7 = 13.4 V.
(a) 50 Hz and 100 Hz; (b) about 14.1 V, or about 13.4 V allowing for the diode
Every mobile charger and LED bulb plugged into India's 50 Hz mains starts with a rectifier that turns AC into DC. A capacitor then smooths it before it reaches the phone battery or the LEDs.
1 marks: “The input to a full-wave rectifier is 50 Hz AC. What is the frequency of its output?” 100 Hz, because there is one output pulse in each half cycle.
3 marks: “Draw the circuit of a half-wave rectifier and explain its working with input and output waveforms.” Transformer secondary, diode and RL in series. Positive half: diode forward biased, output appears. Negative half: reverse biased, no output. Waveforms: sine input; positive halves with gaps.
5 marks: “With a labelled circuit diagram, explain the working of a full-wave rectifier. Draw its input and output waveforms. How can the output be made smoother?” Centre-tap transformer, D1 and D2, RL to the centre tap. D1 conducts in one half, D2 in the other; current through RL is one way in both. Output humps with no gaps at 100 Hz. A capacitor across RL smooths it.
Half uses half, full uses both: one diode gives 50 Hz out; two diodes give 100 Hz out.